Question

Three resistors are connected as shown in the diagram.
Through the resistor $5$ ohm, a current of $1$ ampere is flowing.
(i) What is the current through the other two resistors?
(ii) What is the p.d. across $AB$ and $AC$?
(iii) What is the total resistance?

Answer 1

Given,

Current in AC, $I_{AC}=1A$

Resistance in BC, $R_{BC}=\frac{10\times 15}{10+15}=6\Omega$

Current flowing in series $I_{AB}=I_{BC}=1A$

$V_{BC}=I_{BC}\times R_{BC}=1\times 6=6V$

In parallel potential is equal

$V_{BC}=I_{10}{R}_{10}=I_{15}{R}_{15}=6$

(a) $I_{10}=\frac{6}{10}=0.6A,I_{15}=\frac{0.6}{15}=0.4A$

(b) Potential in AB $V_{AB}=I_{AB}{R}_{AB}=1\times 5=5V$., $V_{BC}=I_{BC}{R}_{BC}=1\times 6=6V$

(c) Total Resistance $R_{total}=R_{AC}+R_{BC}=5\Omega +6\Omega =11\Omega$