Question

(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 $\Omega$.

Answer 1

(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.

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(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}.