Question

In the given circuit, current through 15 Ω resistor will be

• A. $\frac{3}{11}$ A
• B. $\frac{4}{11}$ A
• C. $\frac{5}{11}$ A
• D. Zero

Answer 1

Answer: (B)

$\frac{4}{11}$ A

Answer 2

Let the common node voltage be $V_N$. Each branch gives a current

$$I_1 = \frac{V_N-10}{5},\quad I_2 = \frac{V_N-15}{10},\quad I_3 = \frac{V_N-5}{15}.$$

By KCL at the node,

$$\frac{V_N-10}{5}+\frac{V_N-15}{10}+\frac{V_N-5}{15}=0.$$

Multiplying through by 30 (LCM of 5, 10, 15):

$$6(V_N-10)+3(V_N-15)+2(V_N-5)=0.$$

This simplifies to:

$$6V_N-60+3V_N-45+2V_N-10=11V_N-115=0\quad\Longrightarrow\quad V_N=\frac{115}{11}\,V.$$

Now for the 15 Ω resistor:

$$I_{15}=\frac{V_N-5}{15}=\frac{\frac{115}{11}-5}{15}=\frac{\frac{115-55}{11}}{15}=\frac{60}{11\cdot15}=\frac{4}{11}\;A.$$