Steady State Solution of DC Circuits and Problems based on ohm’s law
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:
RT = R1+R2+R3 (series connection)
IT = VT / RT
V1 = R1*I1
V2= R2*I2
V3 = R3*I3
P1=V1*I1
P2=V2*I2
P3=V3*I3
Solution:
RT = R1+R2+R3 = 2+3+5
RT = 10Ω
IT = VT / RT = 20 / 10
IT = 2 A
In series connection I1 = I2 = I3 = IT = 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 RT = 10Ω
(b). The total current of the circuit IT = 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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