How is impedance Z calculated in series LCR?

$Z = \sqrt{R^2 + (X_L - X_C)^2}$. Here $Z = \sqrt{20^2 + (45-25)^2} = \sqrt{400+400} = \sqrt{800} = 20\sqrt{2} \approx 28.28\ \Omega$.

How is current I calculated?

$I = V/Z = 220/(20\sqrt{2}) = 11/\sqrt{2} \approx 7.78 \approx 7.8$ A.

How is phase angle found?

$\tan\phi = (X_L - X_C)/R = (45-25)/20 = 20/20 = 1$. So $\phi = \tan^{-1}(1) = 45°$.

Is current leading or lagging?

Since $X_L > X_C$, circuit is inductive. Voltage leads current by 45°. Current lags voltage.

What is resonance in LCR circuit?

At resonance $X_L = X_C$, so $Z = R$ (minimum impedance). Current is maximum $I = V/R$. Phase angle $\phi = 0°$ (current and voltage in phase). Resonant frequency $f_0 = 1/(2\pi\sqrt{LC})$.

In a series LCR circuit, $R = 20\ \Omega$, capacitive reactance $X_C = 25\ \Omega$ and inductive reactance $X_L = 45\ \O

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In a series LCR circuit, $R = 20\ \Omega$, capacitive reactance $X_C = 25\ \Omega$ and inductive reactance $X_L = 45\ \Omega$. The circuit is connected to a 220 V AC source. The current in the circuit and the phase angle between current and voltage are respectively:

Options

$11$ A, $\phi = \tan^{-1}(1)$

$7.8$ A, $\phi = 45°$

$11$ A, $\phi = 30°$

$7.8$ A, $\phi = 30°$

Correct Answer

$I = 7.8$ A, $\phi = 45°$

Solution

Net reactance: $X_L - X_C = 45 - 25 = 20\ \Omega$

$$Z = \sqrt{R^2 + (X_L-X_C)^2} = \sqrt{20^2+20^2} = \sqrt{800} = 20\sqrt{2} \approx 28.28\ \Omega$$

$$I = \frac{V}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.8 \text{ A}$$$$\tan\phi = \frac{X_L - X_C}{R} = \frac{20}{20} = 1 \implies \phi = 45°$$

$Z = 20\sqrt{2}\ \Omega$, $I \approx 7.8$ A, $\phi = 45°$

Theory: AC Circuits

1. Series LCR Circuit

In a series LCR circuit connected to AC source $V = V_0\sin(\omega t)$: Resistance $R$: voltage and current in phase. Inductor $X_L = \omega L$: voltage leads current by 90°. Capacitor $X_C = 1/(\omega C)$: current leads voltage by 90°. Impedance: $Z = \sqrt{R^2+(X_L-X_C)^2}$. Current: $I = V/Z$. Phase angle: $\tan\phi = (X_L-X_C)/R$.

2. Resonance Condition

Resonance when $X_L = X_C$: $\omega_0 L = 1/(\omega_0 C)$, so $\omega_0 = 1/\sqrt{LC}$ and $f_0 = 1/(2\pi\sqrt{LC})$. At resonance: $Z_{min} = R$, $I_{max} = V/R$, $\phi = 0$ (unity power factor). $Q$-factor (sharpness) $= \omega_0 L/R = 1/(\omega_0 CR)$. Higher $Q$ = sharper resonance peak.

3. Power in AC Circuits

Average power $P = V_{rms} I_{rms} \cos\phi$. $\cos\phi$ = power factor. Pure resistor: $\cos\phi = 1$, $P = I^2 R$. Pure inductor/capacitor: $\cos\phi = 0$, $P = 0$ (no real power consumed). For this circuit: $P = 220 \times 7.8 \times \cos 45° = 220 \times 7.8 \times 0.707 \approx 1213$ W.

4. Transformer

Works on mutual induction: $V_s/V_p = N_s/N_p = I_p/I_s$. Step-up: $N_s > N_p$ → voltage increases, current decreases. Step-down: $N_s < N_p$ → voltage decreases, current increases. Ideal transformer: $P_{in} = P_{out}$ (100% efficient). Real transformer losses: copper loss ($I^2R$ in windings), iron loss (eddy currents + hysteresis in core). Laminated core reduces eddy currents.

5. Electromagnetic Induction

Faraday law: $\mathcal{E} = -d\Phi/dt$. Lenz law: induced current opposes change causing it. Self-inductance $L$: $\mathcal{E} = -L(dI/dt)$. Mutual inductance $M$: $\mathcal{E}_2 = -M(dI_1/dt)$. Inductance of solenoid: $L = \mu_0 n^2 V$ ($V$ = volume). Energy stored in inductor: $U = \frac{1}{2}LI^2$. Energy stored in capacitor: $U = \frac{1}{2}CV^2$.

6. Alternating Current Basics

$V = V_0\sin(\omega t)$, $I = I_0\sin(\omega t - \phi)$. Peak values $V_0$, $I_0$. RMS values: $V_{rms} = V_0/\sqrt{2}$, $I_{rms} = I_0/\sqrt{2}$. For 220V AC supply: $V_0 = 220\sqrt{2} \approx 311$ V (peak). Frequency in India: 50 Hz, $\omega = 2\pi \times 50 = 100\pi$ rad/s.

7. Phasor Diagram

In series LCR: $V_R$ in phase with $I$. $V_L$ leads $I$ by 90°. $V_C$ lags $I$ by 90°. Net voltage: $V = \sqrt{V_R^2 + (V_L-V_C)^2}$. Phasor diagram shows all voltages as rotating vectors. When $X_L > X_C$: circuit is net inductive, $V$ leads $I$. When $X_C > X_L$: circuit is net capacitive, $I$ leads $V$.

8. Applications of LCR Circuits

Radio tuning: variable capacitor changes $C$ → changes $f_0 = 1/(2\pi\sqrt{LC})$ → selects desired radio station. Filters: low-pass (passes low frequency), high-pass (passes high frequency), band-pass (passes band around $f_0$). Power factor correction: industries use capacitor banks to improve $\cos\phi$ and reduce reactive power. Metal detectors: LCR resonance disturbed by nearby metal → detected.

Frequently Asked Questions

1. Why is $Z = 20\sqrt{2}$ and not just 40? ⌄

$Z = \sqrt{R^2+(X_L-X_C)^2}$. Resistance and reactance add in quadrature (perpendicular in phasor diagram), not directly. $Z = \sqrt{20^2+20^2} = \sqrt{400+400} = \sqrt{800} = 20\sqrt{2} \approx 28.28\ \Omega$. If we simply added: $20+20 = 40\ \Omega$ (WRONG). This is like adding perpendicular sides to get hypotenuse — must use Pythagoras.

2. What is power factor and why does it matter? ⌄

Power factor $\cos\phi$ = ratio of real power to apparent power. Here $\phi = 45°$, so $\cos\phi = 0.707$. Apparent power $S = VI = 220 \times 7.8 = 1716$ VA. Real power $P = S\cos\phi = 1716 \times 0.707 \approx 1213$ W. Reactive power $Q = S\sin\phi \approx 1213$ VAR. Industries pay for real power but supply must deliver apparent power. Low power factor wastes transmission capacity → industries penalised for low power factor.

3. What happens at series resonance? ⌄

At resonance $f_0 = 1/(2\pi\sqrt{LC})$: $X_L = X_C$, net reactance = 0, $Z = R$ (minimum). Current is maximum $= V/R$. Phase angle $\phi = 0$ (unity power factor). Voltage across $L$ and $C$ individually can be much larger than source voltage (they cancel each other). This "voltage magnification" = Q-factor. $V_L = V_C = QV$ at resonance. Dangerous in high-$Q$ circuits (very high voltages across components).

4. How does an inductor behave differently at different frequencies? ⌄

Inductive reactance $X_L = \omega L = 2\pi f L$. At DC ($f = 0$): $X_L = 0$, inductor acts as short circuit (just a wire). At low frequency: small $X_L$, passes current easily. At high frequency: large $X_L$, blocks current. Inductor is a low-pass filter element. Capacitive reactance $X_C = 1/(\omega C) = 1/(2\pi fC)$. At DC: $X_C = \infty$ (blocks DC). At high frequency: $X_C \to 0$ (passes easily). Capacitor is a high-pass filter element.

5. What is Q-factor and how is it calculated here? ⌄

$Q = \omega_0 L/R = X_L/(R)$ at resonance. For this circuit: $Q = X_L/R = 45/20 = 2.25$ (using $X_L$ as proxy for resonance). More precisely, $Q = \omega_0 L/R$ where $\omega_0 = 1/\sqrt{LC}$. High $Q$ means: sharp resonance peak, high selectivity (good for radio tuning), high voltage magnification. $Q = 1/$ bandwidth (in normalised frequency units). $Q > 0.5$: underdamped (oscillations). $Q = 0.5$: critically damped. $Q < 0.5$: overdamped.

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