Question

A charged capacitor is allowed to discharge through a resistance $2\Omega$ by closing the switch $S$ at the instant $t=0$. At time $t=ln2\mu s$, the reading of the ammeter falls half of its initial value. The resistance of the ammeter equal to

• A. $0$
• B. $2\Omega$
• C. $\infty$
• D. $2M\Omega$

Answer 1

Answer: (A)

$0$

For a discharging capacitor the charge on the capacitor with respect to time is given by

$Q=Q_o{e}^{-t/RC}$

So the current through the capacitor can be calculated by differentiation the above equation with respect to time as

$dQ/dt=-\frac{Q_o}{RC}{e}^{-t/RC}$

where the negative signs implies the charge is decreasing from capacitor and hence determines the direction of current. or we can also write

$I=I_o{e}^{-t/RC}$

The resistance R is assumes to be resistance of resistor plus ammeter or

$R=R_{amm}+R_{res}$

Now from above equation of currennt and given information that current falls

half of its initial value at $t=ln2\mu s$, we have

$I_o/2=I_o{e}^{-t/RC}$ or,

$t/RC=ln2$ or,

$R=\frac{t}{Cln2}=\frac{ln2\times {10}^{-6}}{0.5\times {10}^{-6}\times ln2}$ or,

$R=2\Omega$

Now as

$R=R_{res}+R_{amm}$, we have

$2=2+R_{amm}$ or $R_{amm}=0$