Question

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

Answer 1

Correct Answer: 25

We start with the given equation: 

\( \frac{y - 5}{2} + \frac{y - 0}{2} + \frac{y - x + 10}{1} = 0 \)

Expanding and simplifying:

\( y - 5 + y + 2y - 2x + 20 = 0 \)

\( 4y - 2x + 15 = 0 \quad \ldots (1) \)

Next, consider the equation:

\( \frac{x - 5}{4} + \frac{x - 0}{4} + \frac{x - 10 - y}{1} = 0 \)

Expanding this equation:

\( x - 5 + x + 4x - 40 - 4y = 0 \)

\( 6x - 4y - 45 = 0 \quad \ldots (i) \)

Now solving equations (1) and (i):

\( -2x + 4y + 15 = 0 \quad \ldots (ii) \)

and \( 4x - 30 = 0 \)

From these, we get:

\( x = \frac{15}{2} \) and \( 4y - 15 + 15 = 0 \Rightarrow y = 0 \)

Now, the current \( i \) is given by:

\( i = \frac{y - x + 10}{1} \)

Substituting the values of \( x \) and \( y \):

\( i = \frac{0 - 7.5 + 10}{1} \)

\( i = 2.5 \, \text{A} = \frac{n}{10} \, \text{A} \)

Therefore,

\( n = 25 \)

Answer 2

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: \( n = 25 \).