Question

A battery of e.m.f. 15 V and internal resistance $3\Omega$ is connected to two resistors $3\Omega$ and $6\Omega$ connected in parallel. Find:

(a) the current through the battery,

(b) the p.d. between the terminals of the battery,

(c) the current in $3\Omega$ resistor,

(d) the current in $6\Omega$ resistor.

Answer 1

(a) The resultant parallel resistance is given by

$\frac{1}{R}=\frac{1}{3}+\frac{1}{6}\Rightarrow \frac{1}{R}=\frac{1}{2}\Rightarrow R=2\Omega$

Internal resistance, $r=3\Omega$

Current, $I=\frac{E}{R+r}=\frac{15}{2+3}=3A$

(b) P.d. between terminals of battery = Terminal Voltage, $V=E-Ir=15-3\times 3=6V$

(c) Current in 3 $\Omega$ resistor, $I_1=\frac{V}{R_1}=\frac{6}{3}=2A$

(d) Current in 6 $\Omega$ resistor, $I_2=\frac{V}{R_2}=\frac{6}{6}=1A$