Question

An alternating voltage source of peak voltage $V_0$ is connected across a series combination of an inductance L and a variable resistance R as shown in figure. Angle $\varphi$ is the phase difference between voltage and current through source.

• A. For no power dissipation in the circuit $\varphi \rightarrow 0$.
• B. For no power dissipation in the circuit $\varphi \rightarrow \frac{\pi}{2}$
• C. Maximum power dissipation occures for $\varphi \rightarrow \frac{\pi}{4}$

Answer 1

Answer: (B), (C)

Options B and C are correct.

Answer 2

The circuit described is a series L-R (Inductor-Resistor) circuit connected to an alternating voltage source. The instantaneous voltage is $v = V_0 \sin(\omega t)$. The phase difference between the voltage and current is denoted by $\varphi$.

The impedance of the L-R series circuit is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L$ is the inductive reactance. The phase difference $\varphi$ between the voltage and current is given by:

$\tan \varphi = \frac{X_L}{R}$

The average power dissipated in an AC circuit is given by:

$P_{avg} = V_{rms} I_{rms} \cos \varphi$

where $V_{rms} = \frac{V_0}{\sqrt{2}}$ is the RMS voltage and $I_{rms} = \frac{V_{rms}}{Z}$ is the RMS current. The term $\cos \varphi$ is known as the power factor. From the impedance triangle, $\cos \varphi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}$.

Substituting these into the power equation:

$P_{avg} = V_{rms} \left(\frac{V_{rms}}{Z}\right) \left(\frac{R}{Z}\right) = \frac{V_{rms}^2 R}{Z^2} = \frac{V_{rms}^2 R}{R^2 + X_L^2}$

A. For no power dissipation in the circuit $\varphi \rightarrow 0$.

If $\varphi \rightarrow 0$, then $\cos \varphi \rightarrow \cos 0 = 1$. In this case, $P_{avg} = V_{rms} I_{rms} (1) = V_{rms} I_{rms}$. This corresponds to a purely resistive circuit ($X_L = 0$ or $R \rightarrow \infty$). If $X_L = 0$, then $P_{avg} = V_{rms}^2/R$, which is generally non-zero. If $R \rightarrow \infty$, then $I_{rms} \rightarrow 0$, so $P_{avg} \rightarrow 0$. However, $\varphi=0$ generally means maximum power factor, not no power dissipation. Therefore, statement A is incorrect.

B. For no power dissipation in the circuit $\varphi \rightarrow \frac{\pi}{2}$.

For no power dissipation, $P_{avg} = 0$. Since $V_{rms}$ and $I_{rms}$ are generally non-zero, this requires $\cos \varphi = 0$. This implies $\varphi = \frac{\pi}{2}$ (or $90^\circ$). In an L-R series circuit, $\tan \varphi = \frac{X_L}{R}$. If $\varphi = \frac{\pi}{2}$, then $\tan \varphi \rightarrow \infty$, which means $R \rightarrow 0$. When $R=0$, the circuit is purely inductive. In a purely inductive circuit, the current lags the voltage by $\pi/2$, and the average power dissipated is indeed zero (inductors store and release energy, but do not dissipate it as heat). Thus, statement B is correct.

C. Maximum power dissipation occurs for $\varphi \rightarrow \frac{\pi}{4}$.

To find the condition for maximum power dissipation, we need to maximize $P_{avg} = \frac{V_{rms}^2 R}{R^2 + X_L^2}$ with respect to the variable resistance $R$.

We differentiate $P_{avg}$ with respect to $R$ and set the derivative to zero:

$\frac{dP_{avg}}{dR} = V_{rms}^2 \frac{d}{dR} \left( \frac{R}{R^2 + X_L^2} \right)$

Using the quotient rule:

$\frac{dP_{avg}}{dR} = V_{rms}^2 \frac{(R^2 + X_L^2)(1) - R(2R)}{(R^2 + X_L^2)^2} = V_{rms}^2 \frac{R^2 + X_L^2 - 2R^2}{(R^2 + X_L^2)^2} = V_{rms}^2 \frac{X_L^2 - R^2}{(R^2 + X_L^2)^2}$

Setting $\frac{dP_{avg}}{dR} = 0$ gives $X_L^2 - R^2 = 0$, which implies $R^2 = X_L^2$. Since $R$ and $X_L$ are positive, the condition for maximum power dissipation is $R = X_L$.

Now, let's find the phase angle $\varphi$ when $R = X_L$:

$\tan \varphi = \frac{X_L}{R} = \frac{X_L}{X_L} = 1$

Therefore, $\varphi = \tan^{-1}(1) = \frac{\pi}{4}$ (or $45^\circ$). Thus, maximum power dissipation occurs when $\varphi \rightarrow \frac{\pi}{4}$. Statement C is correct.

Both statements B and C are correct based on the principles of AC circuits.

Explanation of the solution:

1. Average Power Formula: The average power dissipated in an AC circuit is $P_{avg} = V_{rms} I_{rms} \cos \varphi$, where $\varphi$ is the phase difference between voltage and current.
2. No Power Dissipation: For $P_{avg} = 0$, we must have $\cos \varphi = 0$, which implies $\varphi = \frac{\pi}{2}$. This occurs when the circuit is purely reactive (e.g., $R=0$ for an L-R circuit).
3. Maximum Power Dissipation: For an L-R series circuit with variable resistance R, the average power is $P_{avg} = \frac{V_{rms}^2 R}{R^2 + X_L^2}$. Differentiating $P_{avg}$ with respect to R and setting to zero yields the condition for maximum power: $R = X_L$.
4. Phase Angle for Maximum Power: When $R = X_L$, the phase angle $\varphi$ is given by $\tan \varphi = \frac{X_L}{R} = \frac{X_L}{X_L} = 1$, which means $\varphi = \frac{\pi}{4}$.