Category: BE8251 Notes

Steady State Solution of DC Circuits and Problems based on ohm’s law

Operating Principles of Moving Iron Instruments Ammeters and Voltmeters

BE8251 Notes r2017 notes

Damping Torque

Damping Torque : We have already seen that the moving system of the instrument will tend to move under the action of the deflecting torque.

But on account of the control torque, it will try to occupy a position of rest when the two torques are equal and opposite.

However, due to inertia of the moving system, the pointer will not come to rest immediately but oscillate about its final deflected position as shown in Fig and takes appreciable time to come to steady state.

To overcome this difficulty a damping torque is to be developed by using a damping device attached to the moving system.

The damping torque is proportional to the speed of rotation of the moving system, that is Tv = kv d dt θ

where kv = damping torque constant

d dt θ = speed of rotation of the moving system

Depending upon the degree of damping introduced in the moving system, the instrument may have any one of the following conditions as depicted in Fig.

Under damped condition: The response is oscillatory
Over damped condition: The response is sluggish and it rises very slowly from its zero position to final position.
Critically damped condition: When the response settles quickly without any oscillation, the system is said to be critically damped.

In practice, the best response is slightly obtained when the damping is below the critical value i.e., the instrument is slightly under damped.

The damping torque is produced by the following methods: Air Friction Damping & Fluid friction damping

Air Friction Damping

In this type of damping a light vane or vanes having considerable area is attached to the moving system to develop a frictional force opposing the motion by reason of the air they displace.

Two methods of damping by air friction are depicted

The arrangement of a light aluminum vane which moves in a quadrant (sector) shaped air chamber.

The chamber also carries a cover plate at the top.

The vane is mounted on the spindle of the moving system.

The aluminum vane should not touch the air-chamber walls otherwise a serious error in the deflection of the instrument will be introduced.

Now, with the motion, the vane displaces air and thereby a damping force is created on the vane that produces a torque (damping) on the spindle.

When the movement is quicker the damping force is greater; when the spindle is at rest, the damping force is zero.

The arrangement of consists of a light aluminum piston which is attached to the moving system.

This piston moves in a fixed chamber which is closed at one end. Either circular or rectangular chamber may be used.

The clearance (or gap) between the piston and chamber walls should be uniform throughout and as small as possible.

When the piston moves rapidly into the chamber the air in the closed space is compressed and the pressure of air thus developed opposes the motion of the piston and thereby the whole moving system.

If the piston is moving out of the chamber, rapidly, the pressure in the closed space falls and the pressure on the open side of the piston is greater than that on the opposite side.

Motion is thus again opposed.

With this damping system care must be taken to ensure that the arm carrying the piston should not touch the sides of the chamber during its movement.

The friction which otherwise would occur may introduce a serious error in the deflection.

The air friction damping is very simple and cheap.

But care must be taken to ensure that the piston is not bent or twisted.

This method is used in moving iron and hot wire instruments.

Fluid Friction Damping

This form is damping is similar to air friction damping.

The action is the same as in the air friction damping.

Mineral oil is used in place of air and as the viscosity of oil is greater, the damping force is also much greater.

The vane attached to the spindle is arranged to move in the damping oil.

It is rarely used in commercial type instruments.

The oil used must fulfill the following requirements.

It should not evaporate quickly.

It should not have any corrosive effect on metals.

Its viscosity should not change appreciably with temperature.

It should be good insulator.

Advantages of Fluid Friction Damping

The oil used for damping can also be used for insulation purpose in some forms of instruments which are submerged in oil.
The clearance between the vanes and oil chamber is not as critical as with the air friction clamping system.
This method is suitable for use with instruments such as electrostatic type where the movement is suspended rather than pivoted.
Due to the up thrust of oil, the loads on bearings or suspension system is reduced thereby the reducing the frictional errors.

Disadvantages of Fluid Friction Damping

The instruments with this type of damping must be kept always in a vertical position.
It is difficult to keep the instrument clean due to leakage of oil.
It is not suitable for portable instruments.
The fluid friction damping can be used for laboratory type electrostatic instruments.

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Steady State Solution of DC Circuits and Problems based on ohm’s law

Electrical Instruments and Classification of instruments

Operating Principles of Moving Iron Instruments Ammeters and Voltmeters

BE8251 Notes r2017 notes

Gravity control

In gravity control – gravity controlled instruments, as shown.

A small adjustable weight is attached to the spindle of the moving system such that the deflecting torque produced by the instrument has to act against the action of gravity.

Thus a controlling torque is obtained.

This weight is called the control weight.

Another adjustable weight is also attached is the moving system for zero adjustment and balancing purpose.

This weight is called Balance weight.

When the control weight is in vertical position as shown

(a), the controlling torque is zero and hence the pointer must read zero.

However, if the deflecting torque lifts the controlling weight from position A to B as shown

(b) such that the spindle rotates by an angle θ, then due to gravity a restoring (or controlling) torque is exterted on the moving system.

The controlling (or restoring) torque, Tc , is given by

Tc = Wl sin θ = k g sin θ where W is the control weight;

l is the distance of the control weight from the axis of rotation of the moving system; and k g is the gravity constant.

Equation shows the controlling torque can be varied quite simply by adjustment of the position of the control weight upon the arm which carries it.

Again, if the deflecting torque is directly proportional to the current,

i.e., Td = kI

We have at the equilibrium position Td = Tc

kI = k g sin θ

I = g k sin θ / k

This relation shows that current I is proportional to sin θ and not θ.

Hence in gravity controlled instruments the scale is not uniform.

It is cramped for the lower readings, instead of being uniformly divided, for the deflecting torque assumed to be directly proportional to the quantity being measured.

Advantanges of Gravity Control

It is cheap and not affected by temperature variations.
It does not deteriorate with time.
It is not subject to fatigue.

Disadvantages of Gravity Control

Since the controlling torque is proportional to the sine of the angle of deflection, the scale is not uniformly divided but cramped at its lower end.
It is not suitable for use in portable instruments (in which spring control is always preferred).
Gravity control instruments must be used in vertical position so that the control weight may operate and also must be leveled otherwise they will give zero error.
In view of these reasons, gravity control is not used for indicating instruments in general and portable instruments in particular.

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Star to Delta transformation

Post author By Mr. Padeepz
Post date January 26, 2018
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Star Delta transformation

The Star Delta transformation are explained in this page.

Star Delta transformation:

Star Delta Transformations allow us to convert impedances connected together from one type of connection to another.

We can now solve simple series, parallel or bridge type resistive networks using Kirchhoff´s Circuit Laws, mesh current analysis or nodal voltage analysis techniques but in a balanced 3-phase circuit

we can use different mathematical techniques to simplify the analysis of the circuit and thereby reduce the amount of math’s involved which in itself is a good thing.

Standard 3-phase circuits or networks take on two major forms with names that represent the way in which the resistances are connected, a Star connected network which has the symbol of the letter, Υ (wye) and a Delta connected network which has the symbol of a triangle, (delta).

If a 3-phase, 3-wire supply or even a 3-phase load is connected in one type of configuration, it can be easily transformed or changed it into an equivalent configuration of the other type by using either the Star Delta Transformation or Delta Star Transformation process.

A resistive network consisting of three impedances can be connected together to form a T or “Tee” configuration but the network can also be redrawn to form a Star or Υ type network as shown below.

As we have already seen, we can redraw the T resistor network to produce an equivalent Star or Υ type network.

But we can also convert a Pi or π type resistor network into an equivalent Delta or type network as shown below.

Pi-connected and Equivalent Delta Network

Having now defined exactly what is a Star and Delta connected network it is possible to transform the Υ into an equivalent circuit and also to convert a into an equivalent Υ circuitusing a the transformation process.

This process allows us to produce a mathematical relationship between the various resistors giving us a Star Delta Transformation as well as a Delta Star Transformation.

These Circuit Transformations allow us to change the three connected resistances (or impedances) by their equivalents measured between the terminals 1-2, 1-3 or 2-3 for either a star or delta connected circuit.

However, the resulting networks are only equivalent for voltages and currents external to the star or delta networks, as internally the voltages and currents are different but each network will consume the same amount of power and have the same power factor to each other.

The value of the resistor on any one side of the delta, network is the sum of all the two-product combinations of resistors in the star network divide by the star resistor located “directly opposite” the delta resistor being found.

For example, resistor A is given as:

A= (PQ + QR + RP) / R with respect to terminal 3 and resistor B is given as:

B = (PQ + QR + RP) / Q with respect to terminal 2 and resistor C given as:

B = (PQ + QR + RP) / R with respect to terminal 1.

By dividing out each equation by the value of the denominator we end up with three separate transformation formulas that can be used to convert any Delta resistive network into an equivalent star network as given below.

Star Delta Transformation allows us to convert one type of circuit connection into another type in order for us to easily analyze the circuit and star delta transformation techniques can be used for either resistances or impedance’s.

One final point about converting a star resistive network to an equivalent delta network.

If all the resistors in the star network are all equal in value then the resultant resistors in the equivalent delta network will be three times the value of the star resistors and equal, giving: R_DELTA = 3R_STAR

Delta to Star Transformation

Compare the resistances between terminals 1 and 2.

P+Q= A in parallel with (B+C)

P+Q = A(B+C) / A+B+C……………….(1)

Resistance between the terminals 2 and 3.

Q+R = C in parallel with (A+B)

Q+R=C(A+B) / A+B+C……………….(2)

Resistance between the terminals 1 and 3.

P+R = B in parallel with (A+C)

P+R = B(A+C) / A+B+C………………(3)

This now gives us three equations and taking equation 3 from equation 2 gives: P+R-Q-R = (B(A+C)) –( C(A+B) ) / A+B+C

P-Q =(BA + BC – CA – BC) / A+B+C P-Q = BA – CA / (A+B+C)…………….(4)

Then, re-writing Equation 1 will give us:

P+Q = (AB+AC) / A+B+C …………………….(5)

Equ (4) + Equ (5)

P+Q+ P-Q = (AB+AC) / A+B+C + (BA – CA) / A+B+C 2P = (AB+AC+BA-CA) / A+B+C

2P = 2AB / A+B+C P = AB / A+B+C

Then to summarize a little about the above maths, we can now say that resistor P in a Star network can be found as Equation 1 plus (Equation 3 minus Equation 2) or Eq1 + (Eq3 – Eq2).

Similarly, to find resistor Q in a star network, is equation 2 plus the result of equation 1 minus equation 3 or Eq2 + (Eq1 – Eq3) and this gives us the transformation of Q as:

Q = AC / A+B+C

and again, to find resistor R in a Star network, is equation 3 plus the result of equation 2 minus equation 1 or Eq3 + (Eq2 – Eq1) and this gives us the transformation of R as:

R = BC / A+B+C

When converting a delta network into a star network the denominators of all of the transformation formulas are the same:

A + B + C, and which is the sum of ALL the delta resistances.

Then to convert any delta connected network to an equivalent star network

If the three resistors in the delta network are all equal in value then the resultant resistors in the equivalent star network will be equal to one third the value of the delta resistors, giving each branch in the star network as: R_STAR = 1/3R_DELTA

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RL Series Circuit And LR Series Circuit

Post author By Mr. Padeepz
Post date January 26, 2018
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RL Series Circuit

RL Series Circuit And LR Series Circuit

RL Series circuit:

In other words, an Inductor in an electrical circuit opposes the flow of current, ( i ) through it.

While this is perfectly correct, we made the assumption in the tutorial that it was an ideal inductor which had no resistance or capacitance associated with its coil windings.

However, in the real world “ALL” coils whether they are chokes, solenoids, relays or any wound component will always have a certain amount of resistance no matter how small associated with the coils turns of wire being used to make it as the copper wire will have a resistive value.

Then for real world purposes we can consider our simple coil as being an “Inductance”, L in series with a “Resistance”, R.

LR Series Circuit

A LR Series Circuit consists basically of an inductor of inductance L connected in series with a resistor of resistance R.

The resistance R is the DC resistive value of the wire turns or loops that goes into making up the inductors coil

The above LR series circuit is connected across a constant voltage source, (the battery) and a switch.

Assume that the switch, S is open until it is closed at a time t = 0, and then remains permanently closed producing a “step response” type voltage input.

The current, i begins to flow through the circuit but does not rise rapidly to its maximum value of Imax as determined by the ratio of V / R(Ohms Law).

This limiting factor is due to the presence of the self induced emf within the inductor as a result of the growth of magnetic flux, (Lenz’s Law).

After a time the voltage source neutralizes the effect of the self induced emf, the current flow becomes constant and the induced current and field are reduced to zero.

We can use Kirchoffs Voltage Law, ( Kirchoffs Voltage Law, (KVL) to define the individual voltage drops that exist around the circuit and then hopefully use it to give us an expression for the flow of current.

Vt = VR + VL

VR = I*R

VL = i dL / dt

V(t) = I*R + i dL / dt

Since the voltage drop across the resistor, V_R is equal to IxR (Ohms Law), it will have the same exponential growth and shape as the current.

However, the voltage drop across the inductor, V_L will have a value equal to: Ve^(-Rt/L).

Then the voltage across the inductor, V_L will have an initial value equal to the battery voltage at time t = 0 or when the switch is first closed and then decays exponentially to zero as represented in the above curves.

The time required for the current flowing in the LR series circuit to reach its maximum steady state value is equivalent to about 5 time constants or 5τ.

This time constant τ, is measured by τ = L/R, in seconds, were R is the value of the resistor in ohms and L is the value of the inductor in Henries.

This then forms the basis of an RL charging circuit were 5τ can also be thought of as “5 x L/R” or thetransient time of the circuit.

The transient time of any inductive circuit is determined by the relationship between the inductance and the resistance.

For example, for a fixed value resistance the larger the inductance the slower will be the transient time and therefore a longer time constant for the LR series circuit.

Likewise, for a fixed value inductance the smaller the resistance value the longer the transient time.

However, for a fixed value inductance, by increasing the resistance value the transient time and therefore the time constant of the circuit becomes shorter.

This is because as the resistance increases the circuit becomes more and more resistive as the value of the inductance becomes negligible compared to the resistance.

If the value of the resistance is increased sufficiently large compared to the inductance the transient time would effectively be reduced to almost zero.

RC Series circuit:

The fundamental passive linear circuit elements are the resistor (R), capacitor (C) and inductor (L).

These circuit elements can be combined to form an electrical circuit in four distinct ways: the RC circuit, the RL circuit, the LC circuit and the RLC circuit with the abbreviations indicating which components are used.

These circuits exhibit important types of behaviour that are fundamental to analogue electronics. In particular, they are able to act as passive filters.

This article considers the RL circuit in both series and parallel as shown in the diagrams.

In practice, however, capacitors (and RC circuits) are usually preferred to inductors since they can be more easily manufactured and are generally physically smaller, particularly for higher values of components.

Both RC and RL circuits form a single-pole filter.

Depending on whether the reactive element (C or L) is in series with the load, or parallel with the load will dictate whether the filter is low-pass or high-pass.

Frequently RL circuits are used for DC power supplies to RF amplifiers, where the inductor is used to pass DC bias current and block the RF getting back into the power supply.

RLC Series Circuit:

Difference between AC AND DC:

Current that flows continuously in one direction is called direct current .

Alternating current (A.C) is the current that flows in one direction for a brief time then reverses and flows in opposite direction for a similar time.

The source for alternating current is called AC generator or alternator.

Cycle:

One complete set of positive and negative values of an alternating quantity is called cycle.

Frequency:

The number of cycles made by an alternating quantity per second is called frequency. The unit of frequency is Hertz(Hz)

Amplitude or Peak value

The maximum positive or negative value of an alternating quantity is called amplitude or peak value.

Average value:

This is the average of instantaneous values of an alternating quantity over one complete cycle of the wave.

Time period:

The time taken to complete one complete cycle.

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AC Instantaneous Value and RMS Value

Post author By Mr. Padeepz
Post date January 26, 2018
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AC Instantaneous Value and RMS Value

AC Instantaneous Value and RMS Value:

AC Instantaneous Value and RMS Value

Instantaneous Value:

The Instantaneous value of an alternating voltage or current is the value of voltage or current at one particular instant.

The value may be zero if the particular instant is the time in the cycle at which the polarity of the voltage is changing.

It may also be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops increasing and starts decreasing.

There are actually an infinite number of instantaneous values between zero and the peak value.

RMS Value:

The average value of an AC waveform is NOT the same value as that for a DC waveforms average value.

This is because the AC waveform is constantly changing with time and the heating effect given by the formula ( P = I ².R ), will also be changing producing a positive power consumption.

The equivalent average value for an alternating current system that provides the same power to the load as a DC equivalent circuit is called the “effective value”.

This effective power in an alternating current system is therefore equal to: ( I².R. Average).

As power is proportional to current squared, the effective current, I will be equal to √ I 2 Ave.

Therefore, the effective current in an AC system is called the Root Mean Squared or RMS.

Pure Resistive circuit:

Resistors are “passive” devices that are they do not produce or consume any electrical energy, but convert electrical energy into heat.

In DC circuits the linear ratio of voltage to current in a resistor is called its resistance.

However, in AC circuits this ratio of voltage to current depends upon the frequency and phase difference or phase angle ( φ ) of the supply.

So when using resistors in AC circuits the term Impedance, symbol Z is the generally used and we can say that DC resistance = AC impedance, R = Z.

It is important to note, that when used in AC circuits, a resistor will always have the same resistive value no matter what the supply frequency from DC to very high frequencies, unlike capacitor and inductors.

For resistors in AC circuits the direction of the current flowing through them has no effect on the behaviour of the resistor so will rise and fall as the voltage rises and falls.

The current and voltage reach maximum, fall through zero and reach minimum at exactly the same time.

i.e, they rise and fall simultaneously and are said to be “in-phase” as shown below.

We can see that at any point along the horizontal axis that the instantaneous voltage and current are in-phase because the current and the voltage reach their maximum values at the same time, that is their phase angle θ is 0^o.

Then these instantaneous values of voltage and current can be compared to give the ohmic value of the resistance simply by using ohms law.

Consider below the circuit consisting of an AC source and a resistor.

The instantaneous voltage across the resistor, V_R is equal to the supply voltage, V_t and is given as:

V_R = V_max sinωt

The instantaneous current flowing in the resistor will therefore be:

I_R = V_R / R

= V_max sinωt / R

= I _max sinωt

In purely resistive series AC circuits, all the voltage drops across the resistors can be added together to find the total circuit voltage as all the voltages are in-phase with each other.

Likewise, in a purely resistive parallel AC circuit, all the individual branch currents can be added together to find the total circuit current because all the branch currents are in-phase with each other.

Since for resistors in AC circuits the phase angle φ between the voltage and the current is zero, then the power factor of the circuit is given as cos 0^o = 1.0.

The power in the circuit at any instant in time can be found by multiplying the voltage and current at that instant.

Then the power (P), consumed by the circuit is given as P = Vrms Ι cos Φ in watt’s. But since cos Φ = 1 in a purely resistive circuit, the power consumed is simply given as, P = Vrms Ι the same as for Ohm’s Law.

This then gives us the “Power” waveform and which is shown below as a series of positive pulses because when the voltage and current are both in their positive half of the cycle the resultant power is positive.

When the voltage and current are both negative, the product of the two negative values gives a positive power pulse.

Then the power dissipated in a purely resistive load fed from an AC rms supply is the same as that for a resistor connected to a DC supply and is given as:

P = V rms * I rms

= I 2 rms * R

= V 2 rms / R

Pure Inductive circuits:

This simple circuit above consists of a pure inductance of L Henries ( H ), connected across a sinusoidal voltage given by the expression: V(t) = V_max sin ωt.

When the switch is closed this sinusoidal voltage will cause a current to flow and rise from zero to its maximum value.

This rise or change in the current will induce a magnetic field within the coil which in turn will oppose or restrict this change in the current.

But before the current has had time to reach its maximum value as it would in a DC circuit, the voltage changes polarity causing the current to change direction.

This change in the other direction once again being delayed by the self-induced back emf in the coil, and in a circuit containing a pure inductance only, the current is delayed by 90^o.

The applied voltage reaches its maximum positive value a quarter ( 1/4ƒ ) of a cycle earlier than the current reaches its maximum positive value, in other words, a voltage applied to a purely inductive circuit “LEADS” the current by a quarter of a cycle or 90^o as shown below.

The instantaneous voltage across the resistor, V_R is equal to the supply voltage, V_t and is given as:

V_L = V_max sin (ωt + 90)

I_L = V / X_L

X_L = 2πfL

Pure Capacitive circuits:

When the switch is closed in the circuit above, a high current will start to flow into the capacitor as there is no charge on the plates at t = 0.

The sinusoidal supply voltage, V is increasing in a positive direction at its maximum rate as it crosses the zero reference axis at an instant in time given as 0^o.

Since the rate of change of the potential difference across the plates is now at its maximum value, the flow of current into the capacitor will also be at its maximum rate as the maximum amount of electrons are moving from one plate to the other.

As the sinusoidal supply voltage reaches its 90^o point on the waveform it begins to slow down and for a very brief instant in time the potential difference across the plates is neither increasing nor decreasing therefore the current decreases to zero as there is no rate of voltage change.

At this 90^opoint the potential difference across the capacitor is at its maximum ( V_max ), no current flows into the capacitor as the capacitor is now fully charged and its plates saturated with electrons.

At the end of this instant in time the supply voltage begins to decrease in a negative direction down towards the zero reference line at 180^o.

Although the supply voltage is still positive in nature the capacitor starts to discharge some of its excess electrons on its plates in an effort to maintain a constant voltage.

These results in the capacitor current flowing in the opposite or negative direction.

When the supply voltage waveform crosses the zero reference axis point at instant 180^o, the rate of change or slope of the sinusoidal supply voltage is at its maximum but in a negative direction, consequently the current flowing into the capacitor is also at its maximum rate at that instant.

Also at this 180^o point the potential difference across the plates is zero as the amount of charge is equally distributed between the two plates.

Then during this first half cycle 0^o to 180^o, the applied voltage reaches its maximum positive value a quarter (1/4ƒ) of a cycle after the current reaches its maximum positive value, in other words, a voltage applied to a purely capacitive circuit “LAGS” the current by a quarter of a cycle or 90^o as shown below.

I_C = I_max sin (ωt + 90)

I_L = V / X_C

X_C = 1 / 2πfC

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Steady State Solution of DC Circuits and Problems based on ohm’s law

Post author By Mr. Padeepz
Post date January 26, 2018
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Steady State Solution of DC Circuits and Problems based on ohm’s law

Resistance in parallel connection:

Resistance in parallel connection: Steady State Solution of DC Circuits and Problems based on ohm’s law

Steady State Solution of DC Circuits:

Resistance in series connection:

The resistors R₁, R₂, R₃ are connected in series across the supply voltage “V”. The total current flowing through the circuit is denoted as “I”. The voltage across the resistor R₁, R₂ and R₃ is V₁, V₂, and V₃ respectively.

V₁ = I*R₁ (as per ohms law)

V₂= I*R₂

V₃ = I*R₃

V = V₁+V₂+V₃

= IR₁+IR₂+IR₃

= (R₁+R₂+R₃) I IR = (R₁+R₂+R₃) I

R = R₁+R₂+R₃

Resistance in parallel connection:

The resistors R₁, R₂, R₃ are connected in parallel across the supply voltage “V”. The total current flowing through the circuit is denoted as “I”. The current flowing through the resistor

R₁, R₂ and R₃ is I₁, I₂, and I₃ respectively.

I = V / R (as per ohms law)

I ₁ = V₁ / R₁

I₂ = V₂ / R₂

I₃ = V₃ / R₃

V₁ = V₂ = V₃ = V

From the above diagram

I = I₁+I₂+I₃

= V₁ / R₁ + V₂ / R₂ + V₃ / R₃

= V / R₁+ V/R₂ +V/R₃

I = V (1/R₁ +1/R₂ +1/R₃)

V / R = V (1/R₁ +1/R₂ +1/R₃)

1/R = 1/R₁ +1/R₂ +1/R₃

Below are problems based on ohm’s law

Problems based on ohm’s law

A current of 0.5 A is flowing through the resistance of 10Ω.Find the potential difference between its ends.

Given data:

Current I= 0.5A.

Resistance R=1Ω

T o f i n d

Potential difference V = ?

Formula used:

V = IR

Solution:

V = 0.5 × 10 = 5V.

Result :

The potential difference between its ends = 5 V

Problems based on ohm’s law

2. A supply voltage of 220V is applied to a 100 Ω resistor. Find the current flowing through it.

Given data

Voltage V = 220V

Resistance R = 100Ω

To find:

Current I = ?

Formula used:

Current I = V / R

Solution:

Current I = 220/100

= 2.2 A

Result:

The current flowing through the resistor = 2.2 A

Problems based on ohm’s law

3. Calculate the resistance of the conductor if a current of 2A flows through it when the potential difference across its ends is 6V.

Given data

Current I = 2A

Voltage V = 6V

To find:

Resistance R = ?

Formula used:

Resistance R = V / I

Solution:

Resistance R = 6 / 2

= 3 Ω

Result:

The value of resistance R = 3Ω

Problems based on ohm’s law

4. Calculate the current and resistance of a 100 W, 200V electric bulb.

Given data:

Power P = 100W

Voltage V = 200V

To find:

Current I =?

Resistance R =?

Formula used:

Power P = V *I

Current I = P / V

Resistance R = V / I

Solution:

Current I = P / V

= 100 / 200

= 0.5 A Resistance R = V / I

= 200 / 0.2

= 400 Ω

Result:

The value of the current I = 0.5 A

The value of the Resistance R = 400 Ω

Problems based on ohm’s law

5. A circuit is made of 0.4 Ω wire, a 150Ω bulb and a 120Ω rheostat connected in series. Determine the total resistance of the circuit.

Given data:

Resistance of the wire = 0.4Ω

Resistance of bulb = 1 5 0 Ω

Resistance of rheostat = 120Ω

To find:

The total resistance of the circuit R _T =?

Formula used:

The total resistance of the circuit R _T = R₁+R₂+R₃

_Solution:

_{Total resistance ,R = 0.4 + 150 +120}

_{= 270.4Ω}

_Result:

The total resistance of the circuit R _T = 270.4 Ω

Problems based on ohm’s law

6. Three resistances of values 2Ω, 3Ω and 5Ω are connected in series across 20 V, D.C supply

.Calculate (a) equivalent resistance of the circuit (b) the total current of the circuit (c) the voltage drop across each resistor and (d) the power dissipated in each resistor.

Given data:

R₁ = 2Ω

R₂ = 3Ω

R₃ = 5Ω

V = 20V

To find:

R _T =?

I _T =?

V₁, V₂, V₃ =?

P₁, P₂, P₃ =?

Formula used:

R_T = R₁+R₂+R₃ (series connection)

I_T = V_T / R_T

V₁ = R₁*I₁

V₂= R₂*I₂

V₃ = R₃*I₃

P₁=V₁*I₁

P₂=V₂*I₂

P₃=V₃*I₃

Solution:

R_T = R₁+R₂+R₃ = 2+3+5

R_T = 10Ω

I_T = V_T / R_T = 20 / 10

I_T = 2 A

In series connection I₁ = I₂ = I₃ = I_T = 2A

V₁ = I₁*R₁ = 2*2

V₁ = 4 V

V₂ = I₂*R₂ = 2*3

V₂ = 6 V

V₃ = I₃*R₃ = 5*2

V₃ = 10V

P₁ = V₁*I₁

= 4*2

P₁ = 8W

P₂ = V₂*I₂

= 6*2

P₂ = 12W

P₃ = V₃*I₃ = 10*2

P₃ = 20W

Result:

(a). Equivalent resistance of the circuit R_T = 10Ω

(b). The total current of the circuit I_T = 2A

(c). Voltage drop across each resistor V₁ = 4 V, V₂ = 6 V, V₃ = 10V

(d). The power dissipated in each resistor P₁ = 8W, P₂ = 12W, P₃ = 20W

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AC Circuits and Kirchhoff’s law

Post author By Mr. Padeepz
Post date January 26, 2018
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Kirchhoff’s Law

AC Circuits and Kirchhoff’s law

AC Circuits and Kirchhoff’s law are explained

AC Circuits

An alternating current (AC) is an electrical current, where the magnitude of the current varies in a cyclical form, as opposed to direct current, where the polarity of the current stays constant.

The usual waveform of an AC circuit is generally that of a sine wave, as this results in the most efficient transmission of energy.

However in certain applications different waveforms are used, such as triangular or square waves

Introduction:

Used generically, AC refers to the form in which electricity is delivered to businesses and residences.

However, audio and radio signals carried on electrical wire are also examples of alternating current.

In these applications, an important goal is often the recovery of information encoded (or modulated) onto the AC signal.

Kirchhoff’s law:

There are Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.

Kirchhoff’s Current Law:

First law (Current law or Point law): Statement:

The sum of the currents flowing towards any junction in an electric circuit equal to the sum of currents flowing away from the junction.

Kirchhoff’s Current law can be stated in words as the sum of all currents flowing into a node is zero.

Conversely, the sum of all currents leaving a node must be zero. As the image below demonstrates, the sum of currents I_b, I_c, and I_d, must equal the total current in I_a.

Current flows through wires much like water flows through pipes.

If you have a definite amount of water entering a closed pipe system, the amount of water that enters the system must equal the amount of water that exists the system.

The number of branching pipes does not change the net volume of water (or current in our case) in the system.

Kirchhoff’s Voltage Law:

Second law (voltage law or Mesh law): Statement:

In any closed circuit or mesh, the algebraic sum of all the electromotive forces and the voltage drops is equal to zero.

Kirchhoff’s voltage law can be stated in words as the sum of all voltage drops and rises in a closed loop equals zero.

As the image below demonstrates, loop 1 and loop 2 are both closed loops within the circuit.

The sum of all voltage drops and rises around loop 1 equals zero, and the sum of all voltage drops and rises in loop 2 must also equal zero.

A closed loop can be defined as any path in which the originating point in the loop is also the ending point for the loop.

No matter how the loop is defined or drawn, the sum of the voltages in the loop must be zero.

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DC Circuits and Ohm’s Law

Post author By Mr. Padeepz
Post date January 26, 2018
No Comments on DC Circuits and Ohm’s Law

DC Circuits and Ohm’s Law Electro-magnetic force(E.M.F) Voltage Potential Difference Electromagnetism Applications of Electromagnetism

DC Circuits and Ohm’s Law

The DC Circuits and Ohm’s Law, Electro-magnetic force(E.M.F), Voltage Potential Difference Electromagnetism Applications of Electromagnetism

DC Circuits

A DC circuit (Direct Current circuit) is an electrical circuit that consists of any combination of constant voltage sources, constant current sources, and resistors.

In this case, the circuit voltages and currents are constant, i.e., independent of time.

More technically, a DC circuit has no memory.

That is, a particular circuit voltage or current does not depend on the past value of any circuit voltage or current.

This implies that the system of equations that represent a DC circuit do not involve integrals or derivatives.

Introduction:

In electronics, it is common to refer to a circuit that is powered by a DC voltage source such as a battery or the output of a DC power supply as a DC circuit even though what is meant is that the circuit is DC powered.

If a capacitor and/or inductor is added to a DC circuit, the resulting circuit is not, strictly speaking, a DC circuit. However, most such circuits have a DC solution.

This solution gives the circuit voltages and currents when the circuit is in DC steady state.

More technically, such a circuit is represented by a system of differential equations.

The solution to these equations usually contains a time varying or transient part as well as constant or steady state part.

It is this steady state part that is the DC solution.

There are some circuits that do not have a DC solution.

Two simple examples are a constant current source connected to a capacitor and a constant voltage source connected to an inductor.

Electro-magnetic force(E.M.F):

Electromotive Force is, the voltage produced by an electric battery or generator in an electrical circuit or, more precisely, the energy supplied by a source of electric power in driving a unit charge around the circuit.

The unit is the volt. A difference in charge between two points in a material can be created by an external energy source such as a battery.

This causes electrons to move so that there is an excess of electrons at one point and a deficiency of electrons at a second point.

This difference in charge is stored as electrical potential energy known as emf.

It is the emf that causes a current to flow through a circuit.

Voltage:

Voltage is electric potential energy per unit charge, measured in joules per coulomb.

It is often referred to as “electric potential”, which then must be distinguished from electric potential energy by noting that the “potential” is a “per-unit-charge” quantity.

Like mechanical potential energy, the zero of potential can be chosen at any point, so the difference in voltage is the quantity which is physically meaningful.

The difference in voltage measured when moving from point A to point B is equal to the work which would have to be done, per unit charge, against the electric field to move the charge from A to B.

Potential Difference:

A quantity related to the amount of energy needed to move an object from one place to another against various types of forces.

The term is most often used as an abbreviation of “electrical potential difference”, but it also occurs in many other branches of physics.

Only changes in potential or potential energy (not the absolute values) can be measured.

Electrical potential difference is the voltage between two points, or the voltage drop transversely over an impedance (from one extremity to another).

It is related to the energy needed to move a unit of electrical charge from one point to the other against the electrostatic field that is present.

The unit of electrical potential difference is the volt (joule per coulomb).

Gravitational potential difference between two points on Earth is related to the energy needed to move a unit mass from one point to the other against the Earth’s gravitational field.

The unit of gravitational potential differences is joules per kilogram.

Electromagnetism:

When current passes through a conductor, magnetic field will be generated around the conductor and the conductor become a magnet.

This phenomenon is called electromagnetism.

Since the magnet is produced electric current, it is called the electromagnet.

An electromagnet is a type of magnet in which the magnetic field is produced by a flow of electric current.

The magnetic field disappears when the current ceases.

In short, when current flow through a conductor, magnetic field will be generated. When the current ceases, the magnetic field disappear.

Applications of Electromagnetism:

Electromagnetism has numerous applications in today’s world of science and physics.

The very basic application of electromagnetism is in the use of motors.

The motor has a switch that continuously switches the polarity of the outside of motor.

An electromagnet does the same thing. We can change the direction by simply reversing the current.

The inside of the motor has an electromagnet, but the current is controlled in such a way that the outside magnet repels it.

Another very useful application of electromagnetism is the “CAT scan machine.”

This machine is usually used in hospitals to diagnose a disease.

As we know that current is present in our body and the stronger the current, the strong is the magnetic field.

This scanning technology is able to pick up the magnetic fields, and it can be easily identified where there is a great amount of electrical activity inside the body

The work of the human brain is based on electromagnetism. Electrical impulses cause the operations inside the brain and it has some magnetic field.

When two magnetic fields cross each other inside the brain, interference occurs which is not healthy for the brain.

Ohm’s Law

Ohm’s law states that the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, and inversely proportional to the resistance between them.

The mathematical equation that describes this relationship is:

I = V/R

where I is the current through the resistance in units of amperes,

V is the potential difference measured across the resistance in units of volts, and R is the resistance of the conductor in units of ohms.

More specifically, Ohm’s law states that the R in this relation is constant, independent of the current.

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