Question

In the circuit shown in figure, the current through the $10\Omega$ resistor is

• A. $\frac{1}{9}A$
• B. $\frac{4}{9}A$
• C. $\frac{2}{3}A$
• D. $\frac{5}{6}A$

Answer 1

The correct option is B $\frac{4}{9}A$

Refer to figure.

Let V be the potential at point E. Points A, B and C are earthed. So the potential at A, B and C = 0. Potential at D = 12 V. Potential difference between D and E is $V_1=12-V$, between E and B is $V_2=V$ and between E and C = V. Therefore
$I_1=\frac{12-V}{3}$
$I_2=\frac{V}{3}$
$I_3=\frac{V}{12}$
From Kirchhoff’s junction rule,
$I_1=I_2+I_3,i.e.$
$\frac{12-V}{3}=\frac{V}{3}+\frac{V}{12}\Rightarrow V=\frac{16}{3}V$
Hence, $I_3=\frac{\frac{16}{3}}{12}=\frac{4}{9}V$