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GUJCET · Physics · Electromagnetic Induction
The Network shown in figure is a part of the circuit. (The battery has negligible resistance).

At a certain instant the current 2 A and it is decreasing at the rate of \(10^2 As ^{-1}\). What is the potential difference between the points \(B\) and \(A\) ?
- A 8.0 V
- B 8.5 V
- C 10 V
- D 15 V
Answer & Solution
Correct Answer
(B) 8.5 V
Step-by-step Solution
Detailed explanation
B
As per second law of Kirchhof
\(\begin{array}{l}
V_{A}-IR+\varepsilon+L \frac{d I}{d t}=V_{B} \\
V_{B}-V_{A}=-IR+\varepsilon+L \frac{d}{d t}
\end{array}\)
\(\begin{aligned} V _{ B }- V _{ A }= & (-2 \times 2)+(12) \\ & +\left(5 \times 10^{-3} \times 10^2\right)\end{aligned}\)
\(\begin{array}{l}V_B-V_A=-4+12+0.5 \\ V_B-V_A=8.5 V\end{array}\)
As per second law of Kirchhof
\(\begin{array}{l}
V_{A}-IR+\varepsilon+L \frac{d I}{d t}=V_{B} \\
V_{B}-V_{A}=-IR+\varepsilon+L \frac{d}{d t}
\end{array}\)
\(\begin{aligned} V _{ B }- V _{ A }= & (-2 \times 2)+(12) \\ & +\left(5 \times 10^{-3} \times 10^2\right)\end{aligned}\)
\(\begin{array}{l}V_B-V_A=-4+12+0.5 \\ V_B-V_A=8.5 V\end{array}\)
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