Question

The network shown in the figure is part of a circuit. If at a certain instant, the current $I$ is $5\text{A}$ flowing through the circuit part and it is decreasing at a rate of ${10}^3{\text{As}}^{-1}$ then $V_B-V_A$ equals

• A. $20\text{V}$
• B. $15\text{V}$
• C. $10\text{V}$
• D. $5\text{V}$

Answer 1

The correct option is B $15\text{V}$

Current $I$ is decreasing, hence emf will be induced across inductor such that it supports the current or will increase the decreasing current.
$\Rightarrow +ve$ terminal of the induced emf will be facing towards $B$.

Now, potential diffrenece across the inductor is given as,
$|\epsilon |=L\frac{di}{dt}=\left(5\text{mH}\right)\times \left({10}^3{\text{As}}^{-1}\right)$
$|\epsilon |=5\text{V}$
$\Rightarrow$Travelling from $A\to B$ and potential will drop across resistance by amount $iR$, potential will increase on crossing battery and inductor.
$V_A-iR+15+L\frac{di}{dt}=V_B$
$V_A-(5\times 1)+15+5=V_B$
$\therefore V_B-V_A=15\text{V}$