Question

The current flowing through the 1\(\Omega \) resistor is \(\frac{n}{10}\)A. The value of n is ________.

Answer 1

Correct Answer: 25

Let the potentials at points A, B, and C be \( x \), \( y \), and \( 0 \), respectively.

Applying Kirchhoff’s Current Law (KCL) at node B:

\[ \frac{y - 5}{2} + \frac{y - 0}{2} + \frac{y - x + 10}{1} = 0 \] \[ \Rightarrow 4y - 2x + 15 = 0 \quad \text{(i)} \]

Applying KCL at node A:

\[ \frac{x - 5}{4} + \frac{x - 0}{4} + \frac{x - 10 - y}{1} = 0 \] \[ \Rightarrow 6x - 4y - 45 = 0 \quad \text{(ii)} \]

Solving equations (i) and (ii):

From (i): \( y = \frac{15}{4}x - \frac{15}{4} \)

Substituting in (ii): \( x = \frac{15}{2}, \, y = 0 \)

The current through the \( 1 \, \Omega \) resistor is:

\[ i = \frac{y - x + 10}{1} = \frac{0 - 7.5 + 10}{1} = 2.5 \, \text{A}. \]

Therefore: \[ i = \frac{n}{10}, \quad n = 25. \]

Final Answer: \( n = 25 \).