Question

For the circuit shown in the following diagram what is the value of [3 MARKS]

(i) current through 6 ohm resistor

(ii) potential difference across 12 ohm

Answer 1

Each point: 1.5 Mark

1. For current through 6 ohm.
Given voltage V= 4 V battery flows through first parallel branch having 6 ohm and 3 ohm in series.
$R=r_1+r_2,R=6+3=9$
Current in this branch.$I=\frac{V}{R},I=\frac{4}{9}=0.44A$

2. For p.d. across 12 ohm.
Given voltage V= 4 V ,Current through second parallel branch. I=?
$R=r_3+r_4,R=12+3=15$
therefore $I=\frac{4}{15}$ A

For p.d across 12 ohm,
$V=IR,V=\frac{4}{15}\times 12=3.2V$