Question

A coil of negligible resistance and inductance $5H$, is connected in series with a $10\Omega$ resistor and a battery of emf $2.0V$. Find the potential difference across the resistor $20ms$ after the circuit is switched on. $(e^{-0.04}=0.096)$

Answer 1

It is an LR circuit.

$L=5H$

$R=10\Omega$

$emf=2V$

$t=20s=2\times {10}^{-2}s$

$i_0=\frac{emf}{R}=\frac{2}{10}=\frac{1}{5}A$

$\tau =\frac{L}{R}=\frac{5}{10}=\frac{1}{2}$

Current equation is given as :

$i=i_0(1-e^{\frac{-t}{\tau }})$

$i=\frac{1}{5}(1-e^{2\times {10}^{-2}\times 2})$

$=\frac{1}{5}(1-e^{-0.04})$

$=\frac{1}{5}(1-0.096)$

$=7.8\times {10}^{-3}A$

$V=iR$

$=7.8\times {10}^{-3}\times 10$

$=0.078V$