(A)Show how you would connect three resistors each of resistance 6 $\Omega$, so that the combination has a resistance of 9 $\Omega$. Also justify your answer. OR (B)In the given circuit, calculate the power consumed in watts in the resistor of 2 $\Omega$.
(A). To achieve a total resistance of \( 9 \, \Omega \), two resistors are connected in parallel, and the third resistor is connected in series with the combination. 1. Parallel combination of two resistors: \[\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6}.\] \[R_p = \frac{6}{2} = 3 \, \Omega.\] 2. Series combination with the third resistor: \[R_{\text{total}} = R_p + R_3 = 3 + 6 = 9 \, \Omega.\] Thus, the required combination is achieved.
(B). The circuit has a \(6 \, \text{V}\) source with resistors \(10 \, \Omega\), \(2 \, \Omega\), and \(10 \, \Omega\) in parallel. The current across the \(2 \, \Omega\) resistor can be calculated using Ohm’s Law. 1. Voltage across each parallel resistor is the same: \[V = 6 \, \text{V}.\] 2. Power consumed in the \(2 \, \Omega\) resistor: \[P = \frac{V^2}{R} = \frac{6^2}{2} = \frac{36}{2} = 18 \, \text{W}.\] Thus, the power consumed in the \(2 \, \Omega\) resistor is {18 watts}.
