01․ The phase angle difference between current and voltage is 90

^{o}, the power will beThe expression of active power P = VIcosθ .
Where, V is voltage, I is current and θ is the angle between current and voltage. Here, this θ = 90

^{o}. ∴ Power P = VIcos90^{o}= 0. (Since, cos90^{o}= 0)02․ Kirchhoffs laws are valid for

Linear circuits obey Ohms law. Kirchhoffs laws are valid for those elements that obey Ohms law.

03․ For the circuit shown below the value of R is adjusted so as to make the current in R

_{L}equal to zero. Calculate the value of R.
As per Wheatstone bridge principle,

04․ In the circuit shown in figure if I

_{1}= 1.5 A, then I_{2}will be
Applying KVL in the loops,
i

_{1}= I_{1}=I_{2}and i_{2}= 0. So I_{2}= 1.5 A.05․ In the circuit shown in the figure the voltage across the 2 Ω resistor is

V

_{B}= 3V Applying KCL at node A, V_{A}= 5V Therefore, the voltage across 2 Ohm resistor is 5-3 = 2V.06․ The value of current I flowing in the 1 Ω resistor in the circuit shown in the given figure will be

5A current flows through the 1 Ohm resistor as no current flows to the voltage source.

07․ The voltage across the 1 Ω resistor between the nodes A and B of the network shown in the given figure is

Applying KCL at node A, V

_{A}= 2V.08․ In the network shown, what is the current I in the direction shown?

V

_{B}= 10V. Applying KCL at node A, V_{A}= 10V. Therefore, the current I is 0.09․ An electrical circuit with 10 branches and 7 nodes will have

Number of loop equations = branches - nodes + 1 = b - n + 1 = 10 - 7 + 1 = 4 loop equations.

10․ 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,
The current through 5 Ω resistor is 8 A hence voltage across it is 8 X 5 = 40 V and this is the voltage across unknown resistor R. As current through R is 2 A.

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