NEET Alternating Current Previous Year Questions with Solutions

Comparing the given equation V = 220 \sin(2 \times 10^3 t) with standard AC form V = V_0 \sin(\omega t):

• • V_0 is the peak voltage amplitude.
• • omega is the angular frequency.

Compute reactances of inductor L and capacitor C:

• • Because X_L = X_C = 20 Omega, the circuit operates at its exact resonant frequency.
• • Net reactance is zero: X_L - X_C = 0.

Impedance of a series LCR circuit:

• • Current amplitude (peak current) is:
• • I_0 = \frac{V_0}{Z} = \frac{220\text{ V}}{100\,\Omega} = 2.2\text{ A}.
• • Matches Option (1) exactly.

In a series LCR circuit, inductive reactance leads the current while capacitive reactance lags by 90°. The net reactance is the algebraic difference:

• • Total impedance Z = √(R² + (X_L - X_C)²) = √(20² + 20²) = 20√2 Ω ≈ 28.28 Ω.

The root-mean-square current supplied by the 220 V AC source is given by Ohm’s law for AC circuits:

• • Current is inductive (voltage leads current by phase angle ϕ).

The phase angle ϕ between voltage and current is determined from the impedance triangle:

• • Thus the current is 7.8 A and the phase angle is 45°, corresponding to Option (2).

In an ideal transformer with negligible winding resistance and zero magnetic flux leakage, the same alternating magnetic flux Φ passes through each turn of both the primary and secondary coils. By Faraday’s law of electromagnetic induction:

Dividing the secondary voltage equation by the primary voltage equation gives the fundamental voltage transformation ratio:

We are given that N_p / N_s = 1 / 2. Inverting gives N_s / N_p = 2 / 1:

The voltage rating given for an AC power supply (210 V) represents its root-mean-square (RMS) voltage. The peak voltage V₀ is related to V_rms by:

For source frequency f = 50 Hz with π = 3.14:

The opposition offered by the capacitor to alternating current is the capacitive reactance:

The peak current I₀ in the pure capacitive circuit is:

Multiplying the values:

Resonance in a series LCR circuit occurs when inductive reactance equals capacitive reactance, causing impedance to become purely resistive (Z = R):

Substitute inductance L = 10 × 10⁻³ H and capacitance C = 10⁻⁶ F:

Convert angular frequency ω₀ (in rad/s) to cyclic frequency f₀ (in Hz):

The opposition offered by a capacitor to alternating current is capacitive reactance X_C:

When the operating frequency f decreases, the capacitive reactance X_C increases inversely:

The alternating current in the circuit (conduction current i_c) is determined by Ohm's law in AC circuits:

Inside the capacitor plates, the changing electric flux generates displacement current i_d, which identically equals the conduction current i_c at every instant:

Since f decreases, i_d decreases proportionally:

In an ideal transformer, energy losses due to flux leakage, eddy currents, hysteresis, and Joule heating in windings are assumed to be zero. Consequently, the input electrical power fed into the primary winding equals the output power delivered by the secondary winding:

The power delivered to the primary winding from the AC supply is given by the product of the primary RMS voltage and primary RMS current (assuming unity power factor for the resistive lamp load):

Substitute the primary mains voltage V_p = 220 V and the secondary load power P_s = 60 W:

Divide both sides by 220 to isolate I_p:

For an AC voltage source operating at a linear frequency of f = 50 Hz, the angular frequency ω is given by:

Substitute inductance L = (50/π) mH = (50/π) × 10⁻³ H into the inductive reactance formula:

Substitute capacitance C = (10³/π) μF = (10⁻³/π) F into the capacitive reactance formula:

In a series LCR circuit, resistance and net reactance combine orthogonally in the impedance triangle:

In a series LCR alternating current circuit, the voltages across the inductor and capacitor are $180^\circ$ out of phase with each other, while the voltage across the resistor is in phase with current. Using the phasor voltage addition theorem:

The problem statement specifies that the current amplitude (peak value $I_0$) is $10\sqrt{2}\text{ A}$. The corresponding RMS current is:

The impedance $Z$ of the LCR series circuit is defined as the ratio of total RMS voltage to total RMS current (or peak voltage $V_0 = 50\sqrt{2}\text{ V}$ to peak current $I_0$):

In an LCR series circuit, power dissipated is $P = I_{\text{rms}}^2 R$. At resonance ($\omega = \omega_0$), power is maximum: $P_{\text{max}} = \frac{V^2}{R}$. The half-power frequencies $\omega_1$ and $\omega_2$ are the frequencies where power drops to half of maximum ($P = P_{\text{max}} / 2$):

Substitute $L = 5.0\text{ H}$ and $C = 80\ \mu\text{F} = 80 \times 10^{-6}\text{ F}$:

Substitute $R = 40\ \Omega$ and $L = 5.0\text{ H}$:

Evaluate the two frequencies:

In an ideal transformer, eddy current, hysteresis, and copper winding losses are completely neglected. By conservation of electrical energy, the power delivered to the primary circuit equals the power consumed by the load in the secondary circuit:

The lamp consumes $P_s = 44\text{ W}$ at primary voltage $V_p = 220\text{ V}$. Rearranging for primary current:

Secondary current is $I_s = \frac{P_s}{V_s} = \frac{44\text{ W}}{11\text{ V}} = 4\text{ A}$. The turns ratio is $\frac{V_s}{V_p} = \frac{11}{220} = \frac{1}{20}$. By current transformation ratio:

When an alternating voltage $$V = V_0 \sin(\omega t)$$ is applied across a pure capacitor of capacitance $$C$$, the opposition to current flow is called capacitive reactance ($$X_C$$): $$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$

By Ohm's law in AC circuits: $$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \cdot (2\pi f C)$$

Given: - $$V_{\text{rms}} = 200\text{ V}$$ - $$f = 50\text{ Hz} \implies \omega = 2\pi \times 50 = 100\pi\text{ rad s}^{-1}$$ - $$C = 40\text{ }\mu\text{F} = 40 \times 10^{-6}\text{ F}$$ $$I_{\text{rms}} = 200 \times (100\pi) \times (40 \times 10^{-6})$$ $$I_{\text{rms}} = 200 \times 4000\pi \times 10^{-6}$$ $$I_{\text{rms}} = 800000\pi \times 10^{-6} = 0.8\pi\text{ A}$$

$$I_{\text{rms}} = 0.8 \times 3.1416 = 2.51328\text{ A} \approx 2.5\text{ A}$$ This matches Option (3).

When $$L$$ is removed, the remaining circuit is a series RC circuit. The phase angle between voltage and current is given by: $$\tan\phi_1 = \frac{X_C}{R}$$ Given $$\phi_1 = \frac{\pi}{3}$$, so: $$\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{X_C}{R} \implies X_C = \sqrt{3} R$$.

When $$C$$ is removed, the remaining circuit is a series RL circuit. The phase angle is: $$\tan\phi_2 = \frac{X_L}{R}$$ Given $$\phi_2 = \frac{\pi}{3}$$, so: $$\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{X_L}{R} \implies X_L = \sqrt{3} R$$.

In the full series LCR circuit, total net reactance is: $$X = X_L - X_C = \sqrt{3}R - \sqrt{3}R = 0$$ Impedance of the circuit is: $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$$. The circuit is at electrical resonance.

The power factor of an AC circuit is defined as: $$\text{Power factor} = \cos\phi = \frac{R}{Z} = \frac{R}{R} = 1.0$$. Matches Option 3.