Power of Alternating Current: Important Questions

Power of alternating current is the ratio of actual power dissipation to the apparent power dissipation in the circuit. The power factor of an AC circuit decreases when a resistor is connected to it whereas it increases when the AC circuit contains an inductor or capacitor connected in series to the resistor. 

Alternating Current Graph

The average value of the power of an AC circuit lies between 0 and 1. In a pure inductive and capacitive circuit, the value is always 0 and for a purely resistive circuit, it is 1. 

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Very Short Answer Questions [1 Mark Questions]

Ques. Write the expression to calculate the average power dissipation in an AC circuit.

Ans. The average power dissipation in an AC circuit is calculated using the formula;

P_av = \(\int_0^t\mathrm{VI \ dt}\ / \int_0^t dt\)

Ques. What is the formula to calculate the apparent power or virtual power of an AC circuit?

Ans. The apparent power of an AC circuit can be calculated with the given formula;

P_av = I²_rms R

Ques. The instantaneous current and voltage of an AC circuit are given by i = 10 sin 314t A and v = 50 sin 314t V. What is the power dissipation in the circuit?

Ans. P = ½ I₀V₀ 

P = (10 \(\times\) 50) / 2 = 250 W

Ques. Define the term 'wattless current’. [Delhi 2011]

Ans. Wattless current is a component of the current in the circuit. It is the current due to which the power consumed in the circuit is zero. 

Ques. A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V DC source? [Delhi 2013]

Ans. I = P/V 

I = 630 / 210 

I = 3 A

Ques. A heating element is marked 210 V, 630 W. Find the resistance of the element when connected to a 210 V DC source.

Ans. Resistance (R) = V2/P 

R = (210)² / 630 

R = 70

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Short Answer Questions [2 Marks Questions]

Ques. Why is the use of AC voltage preferred over DC voltage? 

Ans. AC voltage is preferred over DC voltage due to the following reasons:

1. It can be stepped up or down by a transformer.
2. The carrying losses associated with AC voltage are much less.

Ques. A circuit has a resistance of 12 \(\Omega\) and an impedance of 15\(\Omega\) . What will be the power factor of the circuit?

Ans. Power factor = Cos\(\phi\) = R/Z

Here, Z (impedance) = 15 \(\Omega\)  and R (resistance) = 12 \(\Omega\) 

So, Power factor = 12/15 = 0.8 

Ques. Voltage V and current I in AC circuit is given by V = 50 sin (50t) V and I = 50 sin (50t + \(\pi\)/3) mA. What is the power dissipated in the circuit?

Ans. Power dissipated in an AC circuit (P) = VI cos\(\phi\)

Here, cos \(\phi\) is the power factor and \(\phi\) is the phase angle between voltage and current.

In this question, \(\phi\) = \(\pi\)/3 

So, P = VI cos \(\phi\)

P = 50 0.05 cos(\(\pi\)/3)

P = 1.25 W

Ques. What will be the power dissipated in an AC circuit in which V = 100 sin (100t) V and I = 100 sin (100t + \(\pi\)/3) mA?

Ans. Power (P) = V_rms I_rms cos

P = ½ \(\times\)100\(\times\)(100\(\times\)10^-3) cos (\(\pi\)/3) 

P = 2.5 W

Ques. The instantaneous current and voltage of an AC circuit are given by I = 10 sin 300t A and V = 200 sin 300t V. What is the power dissipation in the circuit?

Ans. P = I_rms\(\times\)V_rms\(\times\)cos\(\phi\)

cos\(\phi\) = 1

P = \((I_0/ \sqrt2 ) \ \times \ (V_0/ \sqrt2)\)

P = ½ I_oV_o 

P = ½ \(\times\)10\(\times\) 200 

P = 1000 W

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Long Answer Questions [3 Marks Questions]

Ques. A light bulb is rated 100 W for 220 V AC supply of 50 Hz. Calculate

(i) the resistance of the bulb

(ii) the rms current of the bulb

Ans. (i). P = V²/R 

R = V²/P = (220)²/100 = 484 \(\Omega\)

(ii) I_rms = V_rms/R 

I_rms = 220/484 = 0.45 A

Ques. An alternating voltage is given by V = 280 sin 50\(\pi\)t is connected across a pure resistor of 40\(\Omega\). Find:

(i) Frequency of the source

(ii) rms current through the resistor 

Ans. (i). V₀ = 140 V, \(\omega\) = 314 

2\(\pi\)v = 314 

v = 314/2\(\pi\) = 50 Hz

(ii) I_rms = V_rms/R

V_rms = V/\(\sqrt{2}\)

I_rms = (V₀/\(\sqrt{2}\))/R 

I_rms = V_o/\(\sqrt{2}\)R = 140 / (\(\sqrt{2}\) * 50) 

I_rms = 140 / (1.414 50)

I_rms = 1.98 A = 2 A

Ques. An inductor of unknown value, a capacitor of 100 \(\mu\)F and a resistor of 10\(\Omega\) are connected in series to a 200V, 50 Hz AC source. It is found that the power factor of the circuit is unity. Calculate the inductance of the inductor and the current amplitude. 

Ans. As power factor of circuit is unity, so

(i) X_L = X_C 

L = 1 / 4\(\pi\)²vc 

L = 1 / (4\(\pi\)²\(\times\)50\(\times\) 100 10-6)

L = 0.1013 H = 101.3 mH

(ii). Current Amplitude,

I_o = V_o/R 

I_o = 2V_rms / R

I_o = (1.414 \(\times\) 200) / 10 

I_o = 28.28 A

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Very Long Answer Questions [5 Marks Questions]

Ques. Two heating elements of resistances R₁ and R₂ when operated at a constant supply of voltage V, consume powers P₁ and P₂ respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in

(i) series

(ii) parallel across the same voltage supply. 

Ans. When two resistances R₁ and R₂ are operated at a constant voltage, the power consumed P₁ and P₂ respectively is,

P₁ = V₂/R₁ and P2 = V²/R² 

When they are connected in series, Power will be,

1/P_s = 1/P₁ + 1/P₂ 

1/P_s = R₁/V² + R₂/V² 

1/P_s = (R₁ + R₂) / V² 

P_s = V² / (R₁ + R₂)

Power in parallel,

P_P = P₁ + P₂ 

P_P = V²/R₁ + V²/R₂ 

P_P = (R₁ + R₂)V² / R₁R₂ 

Ques. For a given AC circuit, i = im sin t, show that the average power dissipated in a resistor R over a complete cycle is ½ i²_mR. 

Ans. i = im sin t, where im is the peak current, is the angular frequency

P = V_rmsI_rmscos\(\phi\)

R = resistance

In AC, across resistance V_rms and I_rms are in phase, cos\(\phi\) = 1

P = V_rmsI_rmscos 0

V_rms = Vm/\(\sqrt2\)

P = V_mi_m / (\(\sqrt2\) \(\times\) \(\sqrt2\))

I_rms = i_m/\(\sqrt2\)

P = V_mi_m / 2

Since, V_m = i_mR 

Hence, P = (i²_m/2)R

Ques. (i). When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero. 

(ii). A Lamp is connected in series with an inductor and an AC source. What happens to the lamp when the key is plugged in and an iron rod is inserted inside the inductor? 

Ans. (i) P_av = I_av\(\times\)e_av cos\(\phi\) = I_AV \(\times\) E_AV cos\(\phi\)

For an ideal inductor, \(\phi\) = \(\pi\)/2 

P_av = I_av eav cos(\(\pi\)/2) 

cos(\(\pi\)/2) = 0

So, P_av = 0 

(ii) When an iron rod is inserted, the value of inductance increases, this leads to decrease in the current and hence brightness decreases. 

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