01․ The phase angle difference between current and voltage is 90o, the power will be
The 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 = 90o. ∴ Power P = VIcos90o = 0. (Since, cos90o = 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 RL equal to zero. Calculate the value of R.
As per Wheatstone bridge principle,
04․ In the circuit shown in figure if I1 = 1.5 A, then I2 will be
Applying KVL in the loops, i1 = I1 =I2 and i2 = 0. So I2 = 1.5 A.
05․ In the circuit shown in the figure the voltage across the 2 Ω resistor is
VB = 3V Applying KCL at node A, VA = 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, VA = 2V.
08․ In the network shown, what is the current I in the direction shown?
VB = 10V. Applying KCL at node A, VA = 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.