Question

For the circuit shown in the figure, find the current passing through $10\Omega$ resistor.

• A. $\frac{5}{3}\text{A}$
• B. $\frac{5}{12}\text{A}$
• C. $\frac{5}{6}\text{A}$
• D. $\frac{5}{18}\text{A}$

Answer 1

The correct option is D $\frac{5}{18}\text{A}$

Let the potential of upper wire be $x$ and the potential of lower wire be zero.



From the figure, we can write that

$i_1+i_2+i_3+i_4+5=10$

$\frac{x}{1}+\frac{x}{2}+\frac{x}{10}+\frac{x}{5}+5=10$

$\Rightarrow x=\frac{50}{18}\text{volts}$

$\therefore i_{10}=\frac{x}{R}=\left(\frac{50}{18}\right)\frac{1}{10}=\frac{5}{18}\text{A}$

Hence, option (a) is the correct answer.

Key concepts : Kirchhoff's Current Law (Junction law) This law is based on law of conservation charge. It states that "The algebraic sum of the currents meeting at a point of the circuit is zero' or total currents entering a junction equals to the total current leaving the junction. $\sum I_{in}=\sum I_{out}$ It is also known as KCL (Kirchoff's current law).