Question

A cell of e.m.f $3V$ and internal resistance $4\Omega$ is connected to two resistance $10\Omega$ and $24\Omega$ joined in parallel. Find the current through each resistance using kirchhoff's laws.

Answer 1

$R_{eq}=4+\frac{10\times 24}{10+24}=11.06\Omega$

$I=\frac{V}{R}=\frac{3}{11.06}=0.27A$

$V_1=V_2$

$I_1{R}_1=I_2{R}_2$

$10I_1=24I_2$

$I_1=2.4I_2$

$I_1+I_2=3.4I_2=0.27$

Current through $24\Omega$ =$I_2=0.08A$

Current through $10\Omega$ =$I_1=0.19A$