(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 obtain a total resistance of \( 9 \, \Omega \), two resistors are connected in parallel, and a third resistor is connected in series with this parallel combination. 1. Parallel connection 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 connection with the third resistor: \[R_{\text{total}} = R_p + R_3 = 3 + 6 = 9 \, \Omega.\] The desired combination is thus achieved.
(B). A \(6 \, \text{V}\) source powers a circuit with three resistors \(10 \, \Omega\), \(2 \, \Omega\), and \(10 \, \Omega\) connected in parallel. Ohm’s Law is used to calculate the current through the \(2 \, \Omega\) resistor. 1. Voltage across each parallel resistor: \[V = 6 \, \text{V}.\] 2. Power consumed by the \(2 \, \Omega\) resistor: \[P = \frac{V^2}{R} = \frac{6^2}{2} = \frac{36}{2} = 18 \, \text{W}.\] Therefore, the power consumed by the \(2 \, \Omega\) resistor is 18 watts.
