JEE Mains · Physics · STD 12 -6. Electromagnetic induction
In the circuit shown here, the point '\(C\)' is kept connected to point '\(A\)' till the current flowing through the circuit becomes constant. Afterward, suddenly, point '\(C\)' is disconnected from point '\(A\)' and connected to point '\(B\)' at time \(t = 0\). Ratio of the voltage across resistance and the inductor at \(t = L/R\) will be equal to:

- A \(1\)
- B \(-1\)
- C \(\frac{{1 - e}}{e}\)
- D \(\;\frac{e}{{1 - e}}\)
Answer & Solution
Correct Answer
(B) \(-1\)
Step-by-step Solution
Detailed explanation
Applying Kirchhof's law of voltage in closed loop \(-V_{R}-V_{C}=0 \Rightarrow \frac{V_{R}}{V_{C}}=-1\)
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