Question

Find the current flowing through the resistance $R_1$ of the circuit shown in the figure if the resistances are equal to $R_1=10\Omega ,R_2=20\Omega ,R_3=30\Omega$ and the potentials of the points 1,2 and 3 are equal to $V_1=10V,V_2=6V$ and $V_3=5V$.

• A. $0.30A$
• B. $0.20A$
• C. $0.40A$
• D. $0.50A$

Answer 1

The correct option is B $0.20A$

Let current in $R_1$ be $I_1$ and $R_2$ be $I_2$ and $R_3$ be $I_3$ respectively away from the junction.

Let junction potential be $V$

By Kirchhoff junction law,

$\frac{V-10}{10}+\frac{V-6}{20}+\frac{V-5}{30}=0$

$\frac{11V}{60}=\frac{7}{6}+\frac{3}{10}=\frac{88}{60}$

$V=\frac{88}{11}$

$V_1=(10-\frac{88}{11})=2$

$I_1=\frac{V_1}{R_1}=\frac{2}{10}=0.2Amp$

Option $\text{B}$ is the correct answer