Q.An alternating current generator has an internal resistance $R_g$ and an internal reactance $X_g$. It is used to supply power to a passive load consisting of a resistance $R_g$ and a reactance $X_L$. For maximum power to be delivered from the generator to the load, the value of $X_L$ is equal to

The key idea is that in an AC circuit, the instantaneous power fluctuates, but the average power depends only on the root-mean-square values of voltage and current and the cosine of the phase angle between them. This is because the product of two sinusoids with a phase shift averages to a constant term proportional to $\cos\phi$, while the oscillating part averages to zero.

1. Identify the phase difference

   Voltage: $V = 0.5 \sin\left(80\pi t + \frac{\pi}{3}\right)$

   Current: $I = 0.5 \sin(80\pi t)$

   The phase of voltage leads the current by $\phi = \frac{\pi}{3}$ radians.

   So the phase difference $\phi = \frac{\pi}{3}$.

2. Recall the formula for average power

   For sinusoidal AC:

where $\cos\phi$ is the power factor.

3. Compute RMS values For a sinusoid of amplitude $A$, $A_{\text{rms}} = \frac{A}{\sqrt{2}}$. Here both voltage and current have amplitude $0.5$:

4. Plug into the formula

The key concept here is power in AC circuits. In an AC circuit, the instantaneous power varies with time, but the meaningful quantity is the average power over a cycle. This average power is given by $P = V_{\text{rms}} I_{\text{rms}} \cos \phi$, where $\phi$ is the phase angle between voltage and current. When $\cos \phi = 0$, the average power is zero — the current is then called "wattless" because it does no net work.

Why does this happen?

• In a purely resistive circuit, voltage and current are in phase ($\phi = 0$), so $\cos \phi = 1$ — maximum power, not wattless.
• In a purely inductive or purely capacitive circuit, voltage and current are exactly 90° out of phase ($\phi = \pm 90^\circ$), so $\cos \phi = 0$ — hence zero average power.
• In an LR or LCR circuit, unless the resistance is zero, there is always some resistive component, so $\cos \phi \neq 0$ and some power is consumed.

Thus, wattless current flows only when the circuit has no resistance and the reactance is purely inductive or purely capacitive. Among the given options, only "purely capacitive circuit" fits (purely inductive is not listed).

Let’s check each option step by step:

1. Option A: LR circuit — Contains resistance $R$ and inductance $L$. The phase angle $\phi$ satisfies $\tan \phi = \frac{\omega L}{R}$, so $\cos \phi \neq 0$ unless $R = 0$ (which is not the case here). Hence power is consumed — not wattless.

2. Option B: Purely capacitive circuit — Only capacitance $C$, no resistance. Voltage lags current by 90° ($\phi = -90^\circ$), so $\cos \phi = 0$. Average power is zero — this is wattless.

3. Option C: Purely resistive circuit — Only resistance $R$. Voltage and current are in phase ($\phi = 0$), so $\cos \phi = 1$. Maximum power is consumed — not wattless.

4. Option D: LCR circuit — Contains resistance, inductance, and capacitance. Unless the circuit is at resonance with $R = 0$ (impossible in practice), $\cos \phi \neq 0$. Even at resonance, if $R > 0$, power is consumed — not wattless.

Concept and Intuition

In AC circuits, the average power is not simply the product of the RMS voltage and RMS current — that product gives the apparent power. The actual power consumed (by a resistor, for example) depends on the phase difference between voltage and current. When they are in phase, all power is real; when out of phase, only the component of current in phase with voltage contributes to average power. That’s why the formula is:

where $\phi$ is the phase angle by which voltage leads current (or current lags voltage). Here, voltage leads current by $\frac{\pi}{6}$ (30°).

Step-by-step solution

1. Identify the RMS values The given current:

The peak current is $I_0 = \frac{3}{\sqrt{2}}$.

RMS current:

Similarly, voltage:

Peak voltage $V_0 = \frac{3}{\sqrt{2}}$, so

2. Find the phase difference

   Current: $\sin(200\pi t)$

   Voltage: $\sin(200\pi t + \pi/6)$

   The voltage leads the current by $\phi = \frac{\pi}{6}$ radians.

3. Compute the power factor

4. Apply the average power formula

Concept and Intuition

In an AC circuit, the average power consumed by a resistive load is given by

where $V_{\text{rms}}$ is the root-mean-square voltage and $I_{\text{rms}}$ is the root-mean-square current. For a sinusoidal voltage $E = E_0 \sin(\omega t + \phi)$, the RMS value is $E_0 / \sqrt{2}$. So the first step is always to convert the given peak voltage to its RMS equivalent. Then, if the total power rating (average power) is known, the RMS current follows directly.

Step-by-step solution

1. Identify the peak voltage

   The voltage is given as $E = 200 \sin(3.14 t + \pi/4)$. The amplitude (peak value) is $E_0 = 200$ V.

2. Compute the RMS voltage

   For a sine wave, $V_{\text{rms}} = \frac{E_0}{\sqrt{2}}$.

3. Relate average power to RMS quantities The total power rating of the devices is given as 500 W. Assuming the load is purely resistive (or that the power factor is unity, which is standard unless otherwise stated),

4. Solve for RMS current

5. Compare with the options

   The computed value $3.54$ A does not match any of the given choices (500 A, 1 A, 200 A, 5 A). This suggests either a misinterpretation or a typo in the problem.

   • If the problem mistakenly treated the peak voltage (200 V) as the RMS voltage, then $I_{\text{rms}} = 500/200 = 2.5$ A — still not an option.
   • If the problem intended the RMS voltage to be 200 V (i.e., the given expression actually represents the RMS value, not the instantaneous), then $I_{\text{rms}} = 500/200 = 2.5$ A — still not an option.
   • The only way to obtain exactly 5 A is if $V_{\text{rms}} = 100$ V, which would correspond to a peak voltage of $100\sqrt{2} \approx 141.4$ V. Since the given peak is 200 V, this is inconsistent.

   Given that 5 A is the only plausible round number among the options (500/100 = 5), the intended answer is most likely (D) 5 A, assuming the problem meant the RMS voltage to be 100 V (or equivalently, the peak voltage to be $100\sqrt{2}$ V).

1. What "ideal" means.

A real transformer loses energy to copper loss ($I^2R$ in the windings), flux leakage, hysteresis of the core, and eddy currents. An ideal transformer is the idealisation in which all of these are zero. Its efficiency is therefore

2. The turns ratio does not change power — it redistributes it.

For an ideal transformer with $N_p$ primary and $N_s$ secondary turns,

Multiplying these two relations:

So whatever the voltage is multiplied by, the current is divided by the same factor. The turns ratio 10 and the 220 V, 50 Hz supply are simply context — they cannot change this conclusion.

3. Why the other options fail.

• (A) and (D) would mean the transformer destroys or creates energy — the latter violates conservation of energy outright.
• (C) zero output power would mean the device does nothing at all.

4. Physical statement.

A transformer is an energy converter, not an energy source. In the ideal limit it hands on every joule it receives.

ANSWER: B

Concept and Intuition

In AC circuits, the average power is not simply the product of the RMS voltage and RMS current — that would give the apparent power. The actual power consumed (the real power) depends on the phase difference between the voltage and the current. When voltage and current are in phase, all power is useful; when they are out of phase, only the component of current that is in phase with voltage contributes to average power. That’s why the formula includes $\cos\phi$, the power factor.

Here, both current and voltage are sinusoidal with the same frequency, but voltage leads current by $\pi/3$ (60°). So we need to:

1. Find the RMS values from the given peak values.
2. Identify the phase difference.
3. Apply the average power formula.

Step-by-step solution

1. Identify the peak values and RMS values The given current:

The peak current is $I_0 = \frac{1}{\sqrt{2}}$ A.

RMS current:

The given voltage:

The peak voltage is $E_0 = \frac{1}{\sqrt{2}}$ V.

RMS voltage:

2. Determine the phase difference Current: phase angle $0$ (reference). Voltage: phase angle $+\pi/3$. So voltage leads current by $\phi = \pi/3$. The power factor is:

3. Apply the average power formula

Substitute:

Concept: series R-L circuit; power factor cos(phi) = R/Z with Z = sqrt(R^2 + XL^2).

From V = 200 sin(100 t): omega = 100 rad/s.

XL = omega L = 100 x 400 x 10^-3 = 40 ohm

Z = sqrt(30^2 + 40^2) = sqrt(900 + 1600) = 50 ohm

cos(phi) = R/Z = 30/50 = 0.6

ANSWER: B

Step 1 — Read off the data.

The ammeter is in series with the source and $R$; the voltmeter is across the series L–C section.

Step 2 — Net reactance.

In a series L–C section the reactances subtract because $V_L$ leads the current by $90^\circ$ while $V_C$ lags it by $90^\circ$:

This is the resonance condition.

Step 3 — Impedance and current.

So the ammeter reads 2 A.

Step 4 — Voltmeter reading across the L–C combination.

The individual magnitudes are equal:

but they are in antiphase. The voltmeter measures their phasor sum:

So the voltmeter reads 0 V, even though each element individually has 8 V across it.

Step 5 — Match to the options.

$0$ V and $2$ A.

ANSWER: B