MCQs on Electric Circuits

01․ In given figure, the value of resistance R in Ω is

The given circuit can be simplified by replacing voltage source by equivalent current source as below,

02․ A 35 V source is connected to a series circuit of 600 Ω and R as shown. If a voltmeter of internal resistance 1.2 kΩ is connected across 600 Ω resistor, it reads 5 V. The value of R is

As the voltmeter of internal resistance 1.2 k Ω is connected across the 600 Ω resistor and it gives 5 V reading. As supply voltage is 35 V, the voltage across resistor R is 35 - 5 = 30 V.

03․ A certain network consists of large number of ideal linear resistances, one of which is designated as R and two constant ideal source. The power consumed by R is P₁ when only the first source is active and P₂ when only second source is active. In both sources are active simultaneously then the power consumed by R is

When first source is active, power P₁ = i₁²R; where i₁ is the current flowing when only the first source is active. When the second source is active, power P₂ = i₂²R; where i₂ is the current flowing when only the second source is active. When both sources are active current flowing through R is

04․ In the circuit given, I = 1 A for I_s = 0. What is the value of I for I_s = 2 A ?

When I_s = 0, V_B = 2V Applying KCL at node A, V_A = 2V Therefore, I = I_s = 2A.

05․ In the circuit shown below, what is the voltage across 5 Ω resistor?

Applying KCL at the nodes, V₁ = -1185.22 V and V₂ = -1156.82 V Therefore the voltage across 5Ohm resistor is -28.4V.

06․ For the circuit shown in the given figure the current I is given by

V_B = 3V Applying KCL at node A, V_A = 5V Therefore, I = (5-3)/2 = 1A

07․ For the circuit given in the figure the power delivered by the 2 V source is given by

Use the superposition theorem. This will give the power delivered by the voltage source without considering the current source. Consider the 2V voltage source with the current source open-circuited. Hence no current pass through the 3 Ohm resistor. So current in the circuit is 2/2 = 1A. So power delivered by the 2V source is V×I = 2×1 = 2W

08․ In the circuit shown in the figure, the value of V_s is 0, when I = 4 A. The value of I, when V_s = 16 V, is

If V_s = 0, then it is short circuited and I = 0.5I_s. So, I_s = 2×4 = 8A. When V_{s = 16 V, Applying KCL at point A, VA = 16V. So, I = VA/2 = 8A.}

09․ Consider the following circuit. In this circuit, when V_s = 3 V, I = 4 A, what is the value of I when V_s = 12 V?

When, V_S = 3V, applying KCl at node A, V_A = 12V and I_S = 7A. When, V_S = 12V, no current will flow through the right hand side branch and I_S = I = 7A.

10․ In the figure given, the value of R is

Applying KCL at node A, V_A = 30V; I₁ = 5A; i₂ = 5A; i₃ = 10A; Also, potential difference between X-Y is 100 - 40 = 60V. Therefore, R = 60/5 = 12 Ohm.