Question

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 4 A and it is increasing at a rate of ${{10}^{4}}$ $A{{s}^{-1}}$ then $({{V}_{p}}-{{V}_{q}})$ is :

A. 56 V
B. 76 V
C. -56 V
D. 66 V

Answer 1

In this question we have been asked to calculate the voltage difference between points P and Q. from the diagram we can say that the given network is a inductor-resistor circuit. Now, from Kirchhoff’s loop rule we know that the sum of all electric potential differences around a loop is zero. Therefore, we shall write the equation for voltage drop and calculate the difference between points P and Q.

Complete step-by-step answer:
Let us assume that the given circuit is a complete loop. Therefore, now we can apply Kirchhoff’s loop rule.
Writing the voltage drop equation for given circuit
We get,
${{V}_{p}}-iR-V-L\dfrac{di}{dt}-{{V}_{q}}=0$
Now, it is given that current I is 4 A. Also, from the diagram we know, inductance L is 5 mH i.e. $5\times {{10}^{3}}$H, voltage V is 10 V and resistance R is 10 Ohms. We have been given that $\dfrac{di}{dt}={{10}^{4}}A{{s}^{-1}}$
Therefore, after substituting the all the given values
We get,
${{V}_{p}}-4\times 4-10-5\times {{10}^{-3}}\times {{10}^{4}}-{{V}_{q}}=0$
On solving
We get,
${{V}_{p}}-{{V}_{q}}=76$

So, the correct answer is “Option B”.

Note: The resistor-inductor circuit consists of a resistor and inductor driven by voltage or current source. The Kirchhoff’s loop rule also known as Kirchhoff’s voltage law states that the sum of voltage difference across a complete loop is always zero. It means that the voltage drop across a loop is zero. This law is similar to conservation of energy in terms of electric potential.