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 P

_{1}when only the first source is active and P_{2}when only second source is active. In both sources are active simultaneously then the power consumed by R isWhen first source is active, power P

_{1}= i_{1}^{2}R; where i_{1}is the current flowing when only the first source is active. When the second source is active, power P_{2}= i_{2}^{2}R; where i_{2}is the current flowing when only the second source is active. When both sources are active current flowing through R is04․ 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, V

_{1}= -1185.22 V and V_{2}= -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 = 1A07․ 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; I_{1}= 5A; i_{2}= 5A; i_{3}= 10A; Also, potential difference between X-Y is 100 - 40 = 60V. Therefore, R = 60/5 = 12 Ohm.<<<4041424344>>>