Q.A sinusoidal voltage of peak value $283\ \text{V}$ and frequency $50\ \text{Hz}$ is applied to a series LCR circuit in which $R = 3\ \Omega$, $L = 25.48\ \text{mH}$, and $C = 796\ \mu\text{F}$. Find

The key idea is what resonance means in a series LCR circuit. When the inductive reactance $X_L = \omega L$ exactly cancels the capacitive reactance $X_C = 1/(\omega C)$, the total impedance becomes purely resistive: $Z = R$. This single fact drives everything — power factor, phase angle, power dissipation, and impedance magnitude.

Let’s check each statement one by one.

1. Power factor becomes one

   Power factor is $\cos \phi$, where $\phi$ is the phase angle between voltage and current. At resonance, $X_L = X_C$, so the net reactance is zero. The impedance is $Z = R$, meaning voltage and current are in phase ($\phi = 0$). Hence $\cos 0 = 1$. Statement (A) is correct.

2. The phase angle between voltages across resistor and source is $90^\circ$

   The voltage across the resistor, $V_R = I R$, is always in phase with the current $I$. The source voltage $V_s$ is also in phase with $I$ at resonance (since $Z = R$). So $V_R$ and $V_s$ are in phase — the phase angle is $0^\circ$, not $90^\circ$. This statement is incorrect.

3. Power dissipation is maximum

   Power dissipated is $P = I^2 R$. At resonance, impedance is minimum ($Z = R$), so current $I = V_s / R$ is maximum for a given source voltage. Hence $P$ is maximum. Statement (C) is correct.

4. Impedance is minimum

   Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$. At resonance, $X_L = X_C$, so $Z = R$, which is the smallest possible value (since any nonzero reactance would make $Z > R$). Statement (D) is correct.

The problem connects three fundamental amplifier parameters: current gain ($\beta$ or $A_i$), power gain ($A_P$), and the input/output resistances. The relationship is not arbitrary — it comes from how power is defined in terms of current and resistance. For a common emitter amplifier, the power delivered to the load is $P_{\text{out}} = I_{\text{out}}^2 R_{\text{out}}$ and the input power is $P_{\text{in}} = I_{\text{in}}^2 R_{\text{in}}$. Taking the ratio gives a clean formula that lets us solve for the unknown output resistance.

1. Write the power gain formula. Power gain is defined as:

But the current gain $A_i$ is $I_{\text{out}} / I_{\text{in}}$, so:

2. Plug in the known values.

   We are given:

   • Current gain $A_i = 50$
   • Power gain $A_P = 3000$
   • Input resistance $R_{\text{in}} = 1200\ \Omega$

   Substituting:

3. Solve for $R_{\text{out}}$. Multiply both sides by 1200:

Divide by 2500:

The core idea here is simple: electrical energy consumed is the product of power and time. Power is the rate at which energy is used, so multiplying by how long the device runs gives the total energy. The unit kilowatt-hour (kWh) is exactly that — energy used by a 1 kW device running for 1 hour. So we just need to convert everything to kilowatts and hours, multiply, and add.

Let’s break it down step by step.

1. Convert power to kilowatts.

   The bulb is $20\ \text{W} = 0.020\ \text{kW}$.

   The fan is $100\ \text{W} = 0.100\ \text{kW}$.

   This conversion is essential because the answer is in kWh.

2. Find daily energy for each device.

   Bulb: $0.020\ \text{kW} \times 5\ \text{h} = 0.10\ \text{kWh}$ per day.

   Fan: $0.100\ \text{kW} \times 15\ \text{h} = 1.50\ \text{kWh}$ per day.

3. Multiply by 30 days to get total energy over the month.

   Bulb: $0.10\ \text{kWh/day} \times 30\ \text{days} = 3\ \text{kWh}$.

   Fan: $1.50\ \text{kWh/day} \times 30\ \text{days} = 45\ \text{kWh}$.

4. Add them up.

   Total = $3\ \text{kWh} + 45\ \text{kWh} = 48\ \text{kWh}$.

The key concept here is phase angle in an AC series LCR circuit. The total opposition to current is impedance $Z = R + j(X_L - X_C)$, where $X_L = \omega L$ and $X_C = 1/(\omega C)$. The phase angle $\phi$ between current and voltage is given by:

• If $\phi > 0$, voltage leads current (inductive behavior).
• If $\phi < 0$, current leads voltage (capacitive behavior).
• If $\phi = 0$, they are in phase (resonance).

So the sign of $X_L - X_C$ directly tells us which leads.

1. Interpret "current leads voltage"

   This means the current reaches its peak before the voltage does. In phasor terms, the current phasor is ahead of the voltage phasor by a positive angle. That implies the phase angle $\phi$ (voltage relative to current) is negative.

2. Relate phase angle to reactances

   From $\tan \phi = (X_L - X_C)/R$, a negative $\phi$ means $\tan \phi < 0$, so $X_L - X_C < 0$. Therefore:

3. Check the options
   • (A) $X_C > X_L$ — matches our result.
   • (B) $X_L > X_C$ — would make voltage lead current.
   • (C) $X_L = X_C \neq 0$ — gives resonance, current and voltage in phase.
   • (D) $X_L = X_C = 0$ — impossible in a real circuit (would mean no inductor or capacitor, but then it's just resistive).

The Core Concept: Phasor Addition in AC Circuits

When you connect a resistor and an inductor in series to an AC source, the current is the same through both components. However, the voltage across the resistor is in phase with the current, while the voltage across the inductor leads the current by $90^\circ$. This phase difference means you cannot simply add the numerical values of the voltages — you must add them as vectors (or phasors), using the Pythagorean theorem.

Think of it like this: if you walk 6 meters east and then 8 meters north, you are not 14 meters from your starting point — you are 10 meters away. The AC voltages behave the same way: the resistor voltage and inductor voltage are perpendicular in phase space.

Step-by-Step Reasoning

1. Identify the given quantities

   The AC source voltage is $V_{\text{source}} = 10\,\text{V}$ (this is the RMS value, as is standard for such problems). The voltage across the inductor is $V_L = 6\,\text{V}$. We need $V_R$, the voltage across the resistor.

2. Recall the phasor relationship

   For a series RL circuit, the source voltage is the vector sum of the resistor voltage and the inductor voltage:

This is because $V_R$ and $V_L$ are $90^\circ$ out of phase.

3. Substitute the known values

4. Solve for $V_R$ Square both sides:

5. Interpret the result The resistor voltage is $8\,\text{V}$. Notice that $6^2 + 8^2 = 36 + 64 = 100 = 10^2$, confirming the phasor relationship.

Concept & Intuition

A capacitor stores charge, and the voltage across it is directly proportional to the charge it holds: $V = Q/C$. Here, the charging current is constant, so charge accumulates at a steady rate: $Q = I t$. The problem asks for the time needed to reach a specific voltage — that’s just the time to accumulate the required charge. No complicated RC time constants; it’s a simple linear relationship.

Step-by-step reasoning

1. Relate charge, capacitance, and voltage For any capacitor, $Q = C V$. We want $V = 100$ V and $C = 100\ \mu\mathrm{F} = 100 \times 10^{-6}\ \mathrm{F}$. So the required charge is

2. Relate charge to constant current A steady charging current $I = 100\ \mu\mathrm{C/s} = 100 \times 10^{-6}\ \mathrm{C/s}$ means charge increases linearly: $Q = I t$. Set this equal to the required charge:

3. Solve for time

Concept & Intuition

A silicon diode in forward bias behaves like a small battery (its forward voltage drop $V_\gamma \approx 0.7$ V) in series with a small internal resistance $r_f$ (here 6 Ω). The load resistor $R_L = 380$ Ω is in series with the diode. The AC source has a peak voltage $V_m = 20$ V. During the positive half-cycle, the diode conducts, and the total voltage available to push current through the circuit is the peak source voltage minus the diode’s fixed drop. The total resistance is the sum of the diode’s internal resistance and the load resistance. Using Ohm’s law gives the peak current; multiplying that current by the load resistance gives the peak load voltage.

Step-by-step solution

1. Identify the circuit elements in series

   The AC source (peak 20 V), the silicon diode (forward drop $V_\gamma = 0.7$ V, internal resistance $r_f = 6\ \Omega$), and the load resistor $R_L = 380\ \Omega$ are all in series.

2. Apply Kirchhoff’s voltage law for the peak of the positive half-cycle

   At the instant the source reaches its positive peak $V_m = 20$ V, the diode is forward-biased. The net voltage driving current is the source voltage minus the diode’s fixed drop:

3. Find the total series resistance The diode’s internal resistance and the load resistor are in series:

4. Calculate the peak diode current Using Ohm’s law:

(More precisely, $19.3 / 386 = 0.05$ exactly, because $19.3 = 386 \times 0.05$.)

5. Calculate the peak load voltage The load voltage is the current through the load resistor times its resistance:

6. Match with the options The pair (50 mA; 19 V) corresponds to option (B). But wait — check the numbers: 50 mA and 19 V appear in both (B) and (C). Option (C) says 52 mA; 19 V. Our calculation gave exactly 50 mA. However, a common pitfall is forgetting the diode’s internal resistance. Let’s verify: if one mistakenly uses only $R_L$ (380 Ω) and neglects $r_f$, the current would be $19.3/380 \approx 50.8$ mA, rounding to 51–52 mA. That matches option (C). The problem explicitly gives the forward junction resistance (6 Ω), so it must be included. With $r_f$ included, the current is exactly 50 mA. Therefore the correct pair is 50 mA and 19 V.