Question

For the circuit shown in the following diagram. What is the value of
(i) Current through $6\Omega$ resistor?
(ii) P.D. across $12\Omega$ resistor ?

Answer 1

(i) $R=6+3=9\Omega$

$V=4V$

$I=\frac{V}{R}=\frac{4}{9}=0.45A$

(ii) 12 ohm and 3 ohm are in series

$R=12+3=15\Omega$

$I=\frac{V}{R}=\frac{4}{15}=0.27A$

p.d across 12 ohms= $IR=0.27\times 12=3.24V$