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

When we consider the 2V voltage source the current source is open circuited and 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.