40. Maximum Power in Variable R Circuit
Step 1: Circuit Impedance and Current
The circuit contains a capacitor $C$ and a variable resistor $R$ in series with an AC source $V = V_0 \sin(\omega t)$.
The RMS voltage is $V_{rms} = V_0 / \sqrt{2}$.
The impedance of the circuit is:
$$ Z = \sqrt{R^2 + X_C^2} \quad \text{where} \quad X_C = \frac{1}{\omega C} $$The RMS current flowing through the load $R$ is:
$$ I_{rms} = \frac{V_{rms}}{Z} = \frac{V_0}{\sqrt{2} \sqrt{R^2 + X_C^2}} $$Step 2: Power Dissipation
The power dissipated in the load resistance $R$ is given by $P = I_{rms}^2 R$:
$$ P(R) = \left( \frac{V_0^2}{2(R^2 + X_C^2)} \right) R = \frac{V_0^2}{2} \left( \frac{R}{R^2 + X_C^2} \right) $$Step 3: Maximizing Power
To find the maximum power, we differentiate $P$ with respect to the variable $R$ and set it to zero ($\frac{dP}{dR} = 0$). This maximizes the term $\frac{R}{R^2 + X_C^2}$.
$$ \frac{d}{dR} \left( \frac{R}{R^2 + X_C^2} \right) = \frac{(R^2 + X_C^2)(1) – R(2R)}{(R^2 + X_C^2)^2} = 0 $$ $$ R^2 + X_C^2 – 2R^2 = 0 \implies R^2 = X_C^2 \implies R = X_C $$
Condition for Max Power: The resistance must equal the capacitive reactance ($R = 1/\omega C$).
Step 4: Calculate Maximum Power
Substitute $R = X_C$ into the power equation:
$$ P_{max} = \frac{V_0^2}{2} \left( \frac{R}{R^2 + R^2} \right) = \frac{V_0^2}{2} \left( \frac{R}{2R^2} \right) = \frac{V_0^2}{4R} $$
$$ P_{max} = \frac{V_0^2}{4R} $$