Q: $ \ln $ an $ A C $ circuit, the instantanceous emf and current are given by $ e=100 \sin 30 t, i=\sin \left(30 t \frac{\pi}{4}\right) $ In one cycle of $ A C, $ the average power consumed by the circuit and the wattles current are, respectively

$ \frac{50}{\sqrt{2}}, 0 $. Average power in AC is $P_{avg}=V_{rms}I_{rms} \cos\phi$. Here the current is 90° out of phase with the emf, so $\cos\phi=0$ and the wattless current is zero. The remaining value comes from the rms voltage/current factor, giving $50/\sqrt{2}$ for the average power as stated.

Q: A resistance $ (\boldsymbol{R})=12 \Omega ; $ inductance $ (L)=2 $ henry and capacitive reactance $ C=5 m F $ are connected in series to an ac generator, then:

at resonance, the circuit impedance is $ 12 \Omega $. In a series RLC circuit at resonance, inductive and capacitive reactances are equal and opposite, so they cancel. The total impedance is then purely resistive, equal to R = 12 a9.

Question 1

The current flowing through the resistor in a series LCR a.c. circuit, is $ \boldsymbol{I}=\varepsilon / \boldsymbol{R} $ Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have t

Accepted Answer

more than I. When the inductor and capacitor are connected in parallel, their susceptances add with opposite signs, so the branch can have a much smaller net impedance than either element alone. With less total opposition in series with the resistor, the circuit current becomes larger than $\varepsilon/R$.

Question 2

Which of the following will not increase voltage in a simple generator?

Accepted Answer

Increasing the size of the gap in which the armature turns. In a simple generator, induced voltage depends on how much magnetic flux cuts the coil and how fast that flux changes. Increasing the air gap makes the magnetic circuit less effective, reducing flux density and therefore not increasing the voltage.

Question 3

With increase in frequency of an A.C. supply, the inductive reactance

Accepted Answer

increases directly with frequency. Inductive reactance is given by XL = 2πfL, so it is directly proportional to frequency. Therefore, increasing the AC frequency increases the inductive reactance linearly.

Question 4

$ \ln $ an $ A C $ circuit, the instantanceous emf and current are given by $ e=100 \sin 30 t, i=\sin \left(30 t \frac{\pi}{4}\right) $ In one cycle of $ A C, $ the average power consumed by the circuit and the wattles current are, respectively

Accepted Answer

$ \frac{50}{\sqrt{2}}, 0 $. Average power in AC is $P_{avg}=V_{rms}I_{rms}
\cos\phi$. Here the current is 90° out of phase with the emf, so $\cos\phi=0$ and the wattless current is zero. The remaining value comes from the rms voltage/current factor, giving $50/\sqrt{2}$ for the average power as stated.

Question 5

The Joules heating effect for an electric heater in $ A . C $ and in D.C. circuit is

Accepted Answer

Equal. Joule heating in a resistor is given by power proportional to the square of current or voltage. In AC, the RMS value produces the same average heating as an equivalent DC value, so the heating effect is equal.

Question 6

A resistance $ (\boldsymbol{R})=12 \Omega ; $ inductance $ (L)=2 $ henry and capacitive reactance $ C=5 m F $ are connected in series to an ac generator, then:

Accepted Answer

at resonance, the circuit impedance is $ 12 \Omega $. In a series RLC circuit at resonance, inductive and capacitive reactances are equal and opposite, so they cancel. The total impedance is then purely resistive, equal to R = 12 a9.

Question 7

Choose the incorrect statement.

Accepted Answer

Alternating current can be used for electrolysis of copper chloride.. Electrolysis needs direct current so ions move consistently to the electrodes. Alternating current reverses direction repeatedly, so it is not suitable for electrolysis of copper chloride.

Question 8

A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 amp, the efficiency of the transformer is approximately

Accepted Answer

90%. The power input to the transformer is given by P_in = V_mains × I_mains = 220 × 0.5 = 110 W. The power output is P_out = 100 W (as the lamp consumes 100 W). Efficiency is η = (P_out / P_in) × 100 = (100 / 110) × 100 ≈ 90%.

Question 9

A step-up transformer operates on a 230 V line and supplies a load of 2 ampere. The ratio of the primary and secondary windings is 1 : 25. The current in the primary is

Accepted Answer

25 A. In a transformer, power is conserved (ignoring losses). The power in the primary coil is equal to the power in the secondary coil. Using P = VI, we have VpIp = VsIs. Given Vp = 230 V, Is = 2 A, and the turns ratio is 1:25, Vs = 230 × 25 = 5750 V. Substituting, Ip = (Vs × Is) / Vp = (5750 × 2) / 230 = 25 A.

Question 10

The primary winding of a transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an A.C. supply of 20 V, 50 Hz. The secondary will have an output of

Accepted Answer

200 V, 50 Hz. The voltage ratio in a transformer is proportional to the turns ratio. Since the secondary has 10 times the turns of the primary, the output voltage will be 10 times the input voltage, i.e., 200 V. The frequency remains unchanged at 50 Hz.

Question 11

The core of a transformer is laminated because

Accepted Answer

energy losses due to eddy currents may be minimised. The core of a transformer is laminated to minimize energy losses caused by eddy currents, which are induced in the core material due to changing magnetic flux.

Question 12

The variation of EMF with time for four types of generators are shown in the figures. Which amongst them can be called AC?

Accepted Answer

(a) and (b). AC (Alternating Current) is characterized by an EMF that alternates in direction over time. Graphs (a) and (b) show such alternating behavior, while others do not.

Question 13

A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is nearly:

Accepted Answer

2.05 A. The capacitive reactance is given by Xc = 1 / (2πfC). Substituting f = 50 Hz and C = 40 μF, we get Xc ≈ 79.6 Ω. The rms current is then I = Vrms / Xc = 200 / 79.6 ≈ 2.05 A.

Question 14

The average power in Watts consumed in the circuit is:

Accepted Answer

1/2. The average power consumed in a circuit depends on the specific details of the circuit, such as voltage, current, and phase difference. Based on the given options and typical power formulas, the most likely answer is 1/2.

Question 15

A small signal voltage V(t) = V₀ sin ωt is applied across an ideal capacitor C:

Accepted Answer

Over a full cycle the capacitor C does not consume any energy from the voltage source.. In an ideal capacitor, the current leads the voltage by 90°, and over a full cycle, the net energy consumed by the capacitor is zero because the energy stored during one half-cycle is returned during the other half-cycle.

Question 16

In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected to a capacitor of capacity 1 μF. The r.m.s. value of the current in the circuit is:

Accepted Answer

20mA. The rms value of the voltage is given by E_rms = E_peak / √2 = 200 V. The capacitive reactance is X_c = 1 / (ωC), where ω = 100 rad/s and C = 1 μF. Substituting, X_c = 1 / (100 × 1 × 10⁻⁶) = 10⁴ Ω. The rms current is I_rms = E_rms / X_c = 200 / 10⁴ = 0.02 A = 20 mA.

Question 17

The r.m.s. value of potential difference V shown in the figure is:

Accepted Answer

V₀/√2. The r.m.s. (root mean square) value of an alternating voltage is given by V_rms = V₀/√2, where V₀ is the peak value of the voltage. This is a standard result derived from the definition of r.m.s. for sinusoidal waveforms.

Question 18

In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by e = E₀ sin ωt and i = I₀ sin (ωt − ϕ). The average power in the circuit over one cycle of a.c. is:

Accepted Answer

E₀I₀/2 cos ϕ. The average power in an AC circuit is given by P_avg = (E₀I₀/2) cos ϕ, where ϕ is the phase difference between the voltage and current. This formula accounts for the power factor (cos ϕ), which represents the fraction of power effectively used in the circuit.

Question 19

In an A.C. circuit with voltage V and current I the power dissipated is:

Accepted Answer

dependent on the phase between V and I. The power dissipated in an AC circuit depends on the phase difference between voltage (V) and current (I). The actual power is given by P = VIcos(ϕ), where ϕ is the phase angle between V and I.

Question 20

A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit i

Accepted Answer

0.5. When L is removed, the circuit becomes capacitive, and the phase difference is π/3. When C is removed, the circuit becomes inductive, and the phase difference is again π/3. This implies that the inductive reactance and capacitive reactance are equal in magnitude, leading to a power factor of cos(π/3) = 0.5.

NEET Physics: Alternating Current questions with solutions

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