Question:
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 :
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 :
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After switching to B the battery is out of the loop, so R and L form a closed loop and the voltages across them must sum to zero.
Updated On: Nov 11, 2026
- \(\frac{e}{1-e}\)
- 1
- -1
- \(\frac{1-e}{e}\)
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The Correct Option is C
Solution and Explanation
With C connected to A the current reaches the steady value \(I_0=\frac{E}{R}\).
After switching to B the battery is removed and the loop contains only R and L. Kirchhoff's loop rule: \(V_R+V_L=0\), i.e. \(IR+L\frac{dI}{dt}=0\).
Hence \(V_R=-V_L\) at every instant, including \(t=L/R\) (where \(I=I_0/e\) is nonzero).
\(\frac{V_R}{V_L}=-1\).
Answer: (3) -1
After switching to B the battery is removed and the loop contains only R and L. Kirchhoff's loop rule: \(V_R+V_L=0\), i.e. \(IR+L\frac{dI}{dt}=0\).
Hence \(V_R=-V_L\) at every instant, including \(t=L/R\) (where \(I=I_0/e\) is nonzero).
\(\frac{V_R}{V_L}=-1\).
Answer: (3) -1
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