1. Identify the components and their configuration: The circuit consists of:
A \( 4 \, \Omega \) resistor (\(R_1\))
A \( 12 \, \Omega \) resistor (\(R_2\))
A resistor with unknown resistance R
A voltage source V
The \( 4 \, \Omega \) and \( 12 \, \Omega \) resistors are connected in parallel. This parallel combination is connected in series with the resistor R. The total equivalent resistance of the circuit (\(R_{eq}\)) is given as \( 10 \, \Omega \).
2. Calculate the equivalent resistance of the parallel combination: Let \(R_p\) be the equivalent resistance of the parallel combination of \( R_1 = 4 \, \Omega \) and \( R_2 = 12 \, \Omega \). The formula for the equivalent resistance of two parallel resistors is: $$ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} $$ Substituting the values: $$ \frac{1}{R_p} = \frac{1}{4 \, \Omega} + \frac{1}{12 \, \Omega} $$ To add these fractions, find a common denominator, which is 12: $$ \frac{1}{R_p} = \frac{3}{12 \, \Omega} + \frac{1}{12 \, \Omega} = \frac{4}{12 \, \Omega} = \frac{1}{3 \, \Omega} $$ Therefore, the equivalent resistance of the parallel part is: $$ R_p = 3 \, \Omega $$ Alternatively, using the product-over-sum formula: $$ R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{4 \, \Omega \times 12 \, \Omega}{4 \, \Omega + 12 \, \Omega} = \frac{48 \, \Omega^2}{16 \, \Omega} = 3 \, \Omega $$
3. Calculate the total equivalent resistance of the circuit: The parallel combination (\(R_p = 3 \, \Omega\)) is connected in series with the resistor R. The total equivalent resistance (\(R_{eq}\)) is the sum of resistances in series: $$ R_{eq} = R_p + R $$
4. Solve for R: We are given that the total equivalent resistance \( R_{eq} = 10 \, \Omega \). Substituting the known values into the equation from Step 3: $$ 10 \, \Omega = 3 \, \Omega + R $$ Subtract \( 3 \, \Omega \) from both sides to find R: $$ R = 10 \, \Omega - 3 \, \Omega $$ $$ R = 7 \, \Omega $$
5. Conclusion: The value of the unknown resistance R is \( 7 \, \Omega \). This corresponds to option (3).
