Question

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

Answer 1

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: $n = 25$.