The current flowing through the 1\(\Omega \) resistor is \(\frac{n}{10}\)A. The value of n is ________.
Correct Answer: 25
Approach Solution - 1
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,
\( n = 25 \)
Approach Solution -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: \( n = 25 \).
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