Question

A capcacitor of capacity 6μF and intial charge 160μC is connected with a key S and different resistors as shown in figure. Point M is earthed. If key is closed at t = 0; then the current through resistance R (=1Ω) at 6 = 16μs is

• A. $\frac{10}{3e}A$
• B. $\frac{10}{e}A$
• C. $\frac{20}{3e}A$
• D. None of the above

Answer 1

The correct option is C

$\frac{20}{3e}A$

Resistance across capacitor

$R=\frac{4\times 8}{4+8}=\frac{32}{12}\Omega$

${\tau }_C=CR=16\mu s$

$i=i_0{e}^{-t/{\tau }_c}=\left(\frac{160/6}{32/12}\right)e^{-t/{\tau }_c}$

$=10e^{-t/{\tau }_c}$

After

$t={\tau }_C,i=\frac{10}{e}A$

This is distributed in inverse ratio of resistance.

$i_n=\frac{2}{3}\left(\frac{10}{e}\right)A=\frac{20}{3e}A$