Question

The Network shown in Figure is a part of the circuit. (The battery has negligible resistance)
At a certain instant the current $I=2A$ and it is decreasing at the rate of ${10}^2{As}^{-1}$. What is the potential difference between the points $B$ and $A$?

• A. $8.0V$
• B. $8.5V$
• C. $10V$
• D. $15V$

Answer 1

The correct option is B $8.5V$

$2△$

Solution:-

Given:-

R=2$\Omega$, I=2A, V=12V, $L=5\times {10}^{-3}H,\frac{dI}{dt}=-100$

Using Kirchoff's law

$V_A-IR+12-L\frac{dI}{dt}=V_B$

$V_B-V_A=-IR+12-L\frac{dI}{dt}$

$V_B-V_A=-2\left(2\right)+12-\frac{5}{1000}(-100)$

$V_B-V_A=8.5V$