Question

A charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t=0$. At J the instant $t=\left(ln4\right)\mu s$, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to

• A. $0.5\Omega$
• B. $1\Omega$
• C. $2\Omega$
• D. $4\Omega$

Answer 1

The correct option is C $2\Omega$

During discharging, current , $i\left(t\right)=i_0{e}^{-t/RC}$

If $r$ be the resistance of ammeter, net resistance , $R=2+r$

When $t=ln4,i=i_0/2\Rightarrow i_0{e}^{-\left(ln4/RC\right)}=i_0/2$

or $e^{\left(ln4/RC\right)}=2$

take ln on both sides, $ln4/RC=ln2$

or $RC=\frac{ln4}{ln2}=\frac{2ln2}{ln2}=2$

or $R=\frac{2}{C}=2/0.5=4$

or $2+r=4\Rightarrow r=2\Omega$