Question

A total charge $q$ passes through a resistor of resistance $R$. The current decreases such that it becomes half of its previous value, after every time $T$. Find the initial value of the current.

• A. $i_0=\frac{q}{T}\ln 2$
• B. $i_0=\frac{q}{T}\ln 4$
• C. $i_0=\frac{q}{T\ln 2}$
• D. $None$ $of$ $these$

Answer 1

The correct option is A $i_0=\frac{q}{T}\ln 2$

The current decreases such that it becomes half of its previous value, after every time T so,

$i_0,\frac{i_0}{2},\frac{i_0}{4},\frac{i_0}{8},\frac{i_0}{16},.....\frac{i_0}{2^n}$
$T_n=t$ , $n=\frac{t}{T}$, $i=\frac{i_0}{2^{t/T}}$
Therefore,
$q={\int }_0^{q}dq={\int }_0^{\infty }idt={\int }_0^{\infty }\frac{i_0}{2^{t/T}}dt=\frac{i_{0}T}{ln2}$

$q=\frac{i_{o}T}{ln2}$
$\Rightarrow i_0=\frac{q}{T}ln2$ . This is the initial value of the current.