Ratio of reading of ammeter A, at t=∞ to that at t = 0 is (switch S is closed at t = 0 )

At t=∞,i=ε2R At t=0, i=εRR2+RR+R2=3ε4RRatio= ε2R4R3ε=23

Ratio of reading of ammeter A, at t=∞ to that at t = 0 is (switch S is closed at t = 0 )

At t=∞,i=ε2R At t=0, i=εRR2+RR+R2=3ε4RRatio= ε2R4R3ε=23

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Ratio of reading of ammeter A, at $t = \infty$ to that at t = 0 is (switch S is closed at t = 0 )

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$A t t = \infty, i = \frac{ε}{2 R} A t t = 0, i = \frac{ε}{\begin{array}{l} R \frac{R}{2} + R \\ R + \frac{R}{2} \end{array}} = \frac{3 ε}{4 R}$

Ratio= $\frac{ε}{2 R} \frac{4 R}{3 ε} = \frac{2}{3}$

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Ratio of reading of ammeter A, at $t = \infty$ to that at t = 0 is (switch S is closed at t = 0 )