Question

In the circuit below, the two DC voltage sources have voltages of value \(V_1\) and \(V_2\). The expression for the power dissipated in the \(60\,\text{k}\Omega\) resistor is proportional to ____________________.

• A. \((V_1+V_2)^2\)
• B. \((3V_1+V_2)^2\)
• C. \((2V_1+V_2)^2\)
• D. \((V_1+2V_2)^2\)

Hint:

For power across a resistor to ground, first find the node voltage via nodal (superposition) analysis; the power is then proportional to the square of that node voltage.

Answer 1

The Correct Option is C

Step 1: Let the voltage at the node connected to the \(60\,\text{k}\Omega\) resistor (center node) be \(V_B\). By nodal analysis: \[ \frac{V_B - V_1}{10k} + \frac{V_B - V_2}{20k} + \frac{V_B}{60k} = 0 . \] Multiplying by \(60k\) gives \[ 6(V_B - V_1) + 3(V_B - V_2) + V_B = 0 \Rightarrow 10V_B - 6V_1 - 3V_2 = 0. \] Hence, \[ V_B = \frac{6}{10}V_1 + \frac{3}{10}V_2 = 0.3(2V_1 + V_2). \] Step 2: The power in the \(60\,\text{k}\Omega\) resistor is \(P = \dfrac{V_B^2}{60k}\), which is proportional to \(V_B^2\). Therefore, \[ P \propto \big(0.3(2V_1+V_2)\big)^2 \propto (2V_1+V_2)^2. \] Thus, the required proportional expression is \(\boxed{(2V_1+V_2)^2}\).