Physics AC Currents

1 Mark Questions

1.A reactive element in an AC circuit causes the current flowing

1. to lead in phase by <math>\pi/2</math>
2. to lag in phase by <math>\pi/2</math> w.r.t. the applied voltage. Identify the element in each case. [Delhi 2010C]

Ans

1. In case of pure capacitive circuit, the current leads in phase by <math>\pi/2</math> with respect to the applied voltage. So, the element will be a capacitor. (1/2)
2. In case of pure inductive circuit, the current lags in phase by <math>\pi/2</math> with respect to the applied voltage. So, the element will be an inductor. (1/2)

2 Marks Questions

2.A capacitor C, a variable resistor R and a bulb B are connected in series to the AC mains in the circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor keeping resistance R to be the same (ii) the resistance R is increased keeping the same capacitance? [Delhi 2014]

Mains

Ans.

1. As, the dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease, i.e. <math>V = \frac{Q}{C}</math>. As a result, the potential drop across the bulb will increase as they are connected in series. Thus, its brightness will increase. (1)

1. As the resistance R is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease as they are connected in series. Thus, its brightness will decrease. (1)

3. The figure shows a series L-C-R circuit connected to a variable frequency 200 V source with <math>L = 50 \text{ mH}</math>, <math>C = 80 \mu\text{F}</math> and <math>R = 40 \Omega</math>.

Determine

1. the source frequency which derives the circuit in resonance.
2. the quality factor (Q) of the circuit.

[All India 2014C]

<math>\mathcal{N}^R</math>

m Ans.

Given, <math>L = 50 \text{mH} = 50 \times 10^{-3} \text{H}</math>

<math>C = 80 \,\mu\text{F} = 80 \times 10^{-6} \,\text{F}</math>

<math>R = 400\pi</math>, <math>V = 200 \text{ V}</math>

1. In the L-C-R, the resonant angular frequency is given by

<math>\omega_0 = -</math> <math display="block">= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{50 \times 10^{-3} \times 80 \times 10^{-6}}}</math> <math>= 500 \text{rad/s}</math>

... The actual/source frequency is given by

<math>\omega = 2\pi v \implies v = \frac{\omega}{2\pi}</math>

<math>\Rightarrow</math> v = <math>\frac{500}{2\pi} = \frac{250}{\pi} = 79.61 \approx 80 \text{ Hz}</math>

(ii) Quality factor, <math>Q = \frac{\omega_0 L}{R}</math> <math display="block">=\frac{500\times50\times10^{-3}}{40}=0.625</math>

4. The figure shows a series L-C-R circuit connected to a variable frequency 250 V source with <math>L = 40 \,\mathrm{mH}, \, C = 100 \,\mu\mathrm{F} \,\mathrm{and} \, R = 50 \,\Omega.</math>

Determine

1. the source frequency which derives the circuit in resonance,
2. the quality factor (Q) of the circuit. [All India 2014]

₩

<math>\overline{m}</math>

Ans. Refer to ans3