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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:

  1. A \(1\)
  2. B \(-1\)
  3. C \(\frac{{1 - e}}{e}\)
  4. D \(\;\frac{e}{{1 - e}}\)
Verified Solution

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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