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

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; 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>>>