## Karnataka 1st PUC Electronics Question Bank Chapter 5 Application of DC and AC to Passive Components

One Mark Questions and Answers

Question 1.

What is a transient period?

Answer:

The transient period is the time period during which current and voltage changes before a steady state is reached when dc is applied to RC and RL circuit.

Question 2.

What is the transient phenomenon?

Answer:

The transient phenomenon is the voltage and current response with respect to time.

Question 3.

Define the time constant of an RC circuit.

Answer:

The time constant of the RC circuit is the time taken by the capacitor to charge to 63.2% of the supply voltage.

Question 4.

Write the expressions for the (1) Voltage across capacitor during charging and (2) voltage across capacitor during discharging.

Answer:

- During charging, V
_{c}=E(1 – e^{-1/RC}) - During discharging, V
_{c}= Ee^{-1/RC}

Question 5.

Define the time constant of the R_{L} circuit.

Answer:

The time constant of the R_{L} circuit is the time taken by the current to grow to 63.2% of the maximum value.

Question 6.

Define the phase difference between two AC quantities.

Answer:

Two ac quantities are said to have a phase difference if they do not reach their maximum or zero values simultaneously.

Question 7.

What are the values of the phase angle between two AC quantities (i) in phase (ii) out of phase?

Answer:

- 0°
- 180°.

Question 8.

Draw the waveforms of (1) two in-phase AC quantities (2) out of phase AC quantities.

Answer:

Question 9.

Write the expression for the resonance frequency of a series resonance circuit.

Answer:

f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

Question 10.

Define half-power frequencies.

Answer:

Half power frequencies are the frequencies at which power delivered to the circuit is half of the power delivered at resonance.

Question 11.

How are the values of current and impedance in the RLC circuit at resonance?

Answer:

At resonance, the current is maximum and impedance is minimum.

Question 12.

Define bandwidth.

Answer:

Bandwidth is the difference between the two frequency points on either side of the resonant frequency where the current rises to 70.7% of its maximum value.

Question 13.

Define quality factor.

Answer:

The quality factor of a circuit is the ratio of resonant frequency to the bandwidth.

Question 14.

Write the expression for the capacitive reactance of a capacitor.

Answer:

Capacitive reactance X_{c} = \(\frac{1}{2 \pi \mathrm{fc}}\)

Question 15.

Write the expression for the inductive reactance of an inductor.

Answer:

Inductive reactance, X_{L} = 2πfL

Two Mark Questions and Answers

Question 1.

Draw the graph showing voltage across the capacitor during charging.

Answer:

Question 2.

Draw the graph of growth of current in RL circuit.

Answer:

Question 3.

Draw the circuit diagrams of low pass filter and high pass filter.

Answer:

Question 4.

Derive an expression for the resonance frequency of a series resonance circuit.

Answer:

At resonance, X_{c} = X_{L} .

2πfL = \(\frac{1}{2 \pi \mathrm{fL}}\)

f^{2} = \(\frac{1}{4 \pi_{2} \mathrm{LC}}\)

f = \(\frac{1}{2 \pi \sqrt{L C}}\)

Question 5.

What is the impedance of a circuit?

Answer:

The impedance of a circuit is the total opposition offered by the RLC circuit for the flow of alternating current.

Question 6.

Draw phasor diagrams for an ac circuit containing (i) a pure inductor and (ii) a pure capacitor only.

Answer:

Question 7.

Determine the time constant of an RC circuit when the resistor is 20kΩ and the capacitor is 0.05 μF.

Answer:

τ = RC = 20 × 10^{-3} × 0.05 × 10^{-6} = 1 × 10^{-3}s = 1 ms

Question 8.

What is the capacitive reactance of 0.01 μF capacitor at 400 Hz?

Answer:

X_{c} = \(\frac{1}{2 \pi \mathrm{fc}}=\frac{1}{2 \times 3.14 \times 400 \times 0.01 \times 10^{-6}}\) = 39.8kΩ

Question 9.

Time constant of RL circuit is 4 ms. IF L = 100 mil, calculate the value of resistance.

Answer:

τ = \(\frac{\mathrm{L}}{\mathrm{R}}\), τR = L, R = \(\frac{\mathrm{L}}{\tau}=\frac{100 \times 10^{-3}}{4 \times 10^{-3}}\) = 25Ω

Question 10.

What value of resistance must be connected in series with a 20μF capacitor to provide a time constant of 0.2s?

Ans.

τ = RC, R = \(\frac{\tau}{\mathrm{c}}=\frac{0.2}{20 \times 10^{-6}}=\frac{2}{200}\) × 10^{6} = 10,000Ω = 10KΩ

Three Mark Questions and Answers

Question 1.

What is the phase difference between the voltage and current in an ac circuit with

(i) pure resistor only

Answer:

0°

(ii) pure capacitor

Answer:

90°

(iii) pure inductor.

Answer:

90°.

Question 2.

What is capacitive reactance? Mention its S.L unit and write the expression for it.

Answer:

Capacitive reactance of a capacitor is the opposition offered by.it for the flow of ac. Its S.I. unit is Ω capacitive reactance, X_{c} = \(\frac{1}{2 \pi \mathrm{fc}}\)

Question 3.

Define

(i) average power

Answer:

Average power or active power is the average instantaneous power over one cycle.

(ii) Reactive power

Answer:

Reactive power is the maximum power consumed in a reactive element.

(iii) Apparent power.

Answer:

Apparent power is the power drawn from an ac source.

Question 4.

Define quality factor and write the relation between quality factor, bandwidth, and resonant frequency.

Answer:

The quality factor of a circuit is the ratio of resonant frequency to the bandwidth.

Q = \(\frac{\text { Resonant frequency }}{\text { Bandwidth }}=\frac{\mathrm{f}_{\mathrm{r}}}{\mathrm{BW}}\)

Question 5.

A series resonant circuit has R = 100 Ω, C = 0.1 μF produces a resonant frequency of 3kHz. Find the value of inductance.

Answer:

Question 6.

A coil of 100 mH having a resistance of 100Ω is connected across a source of 200 V, 50 HZ. Find the phase angle.

Answer:

Given L = 100mH,= 100 × 10^{-3}H = 0.1H. R = 100Ω V = 200V, f = 50Hz.

V = 200V, f = 50Hz

Q = tan^{-1}\(\left(\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right)\) = tan^{-1}\(\left(\frac{2 \pi \mathrm{fL}}{\mathrm{R}}\right)\)

= tan^{-1}\(\left[\frac{2(3.14)(50) 0.1}{100}\right]\)

= tan^{-1}(0.314)

= 17°43 1

Question 7.

Determine the voltage across capacitor and current during charging at t = 1s in a DC circuit containing R = 1KΩ and C = 1mF connected to DC supply of 10V.

Answer:

E = 10V, t= l.s, R= 1KΩ = 1 × 10^{3}Ω .

C= 1mF = 1 × 10^{-3}F.

Question 8.

8. Determine time constant and peak current of an RL circuit with DC source of 10V having resistance R = 100Ω and L = 10 mH

Answer:

Give E = 10V, R = 100Ω L = 10mH = 10 × 10^{3}H

τ = \(\frac{L}{R}=\frac{10 \times 10^{-3}}{100}\) = 0.1 ms

I = \(\frac{\mathrm{E}}{\mathrm{R}}=\frac{10}{100}\) = 0.1 A

Five Mark Questions and Answers

Question 1.

Discuss the charging of a capacitor in an RC circuit.

Answer:

Consider a resistor R and a capacitor C connected in series with a battery of emf E. When the switch is in position A, the capacitor C gets charged through R exponentially with time.

Hence Voltage across the capacitor increases exponentially and the current in the circuit decreases exponentially. When the capacitor is fully charged, the current becomes Zero. The instantaneous value of voltage across the capacitor during charging is given by

V_{C} = v_{o}(i – e^{t/RC})

The instantaneous value of current is I = I_{o}e^{t/RC} Time constant (τ) of an RC circuit is the time taken by the capacitor to charge to 63.2% of maximum current.

Question 2.

Discuss the discharging of capacitors in an RC circuit.

Answer:

When the switch is in position A, the capacitor gets charged. When the switch is in position B, the capacitor starts getting discharged through the resistor R. The voltage across the capacitor decreases and it takes infinite time to get discharged completely. Discharging voltage across the capacitor at a time after the switch is input in the position B is, V_{c} = V_{o}e^{-t/RC} The time constant can be defined as the time taken by the capacitor to discharge to 36.8% of its maximum voltage.

Question 3.

Discuss the growth of current in the RL circuit.

Answer:

Consider an inductor Land a resistor R connected in series to a cell. When the switch is in position A, the current in the circuit grows exponentially.

The current I at a time after the switch is placed in position A is I = I_{o}e^{t/RC}, where lo is the maximum current.

The time constant of RL circuits is the time taken by the current to grow to 63.2% of the maximum value.

Question 4.

Discuss decay of current in RL circuit.

Answer:

When the switch is in position A, the current in the circuit increases exponentially and reaches a maximum value. When the switch is in position B, the current in the circuit starts decaying exponentially with time.

The value of current at a time after the switch is placed in position B is I = I_{o}e^{(-R/L)t} where I_{o} is the maximum current.

The time constant of the RL circuit is the time taken by the RL circuit for the current to decay to 36.8% of its maximum value.

Question 5.

Describe the phenomenon of resonance in a series resonant circuit.

Answer:

Consider an AC circuit consisting of inductance (L), capacitor C and a resistance R connected in series to an AC source. Let the instantaneous voltage of the ac source be

v = V_{m} sin wt,

Let V_{L}, V_{R}, V_{C} be the voltages across the inductor, Resistor, and capacitor.

Result Voltage V = \(\sqrt{V_{\mathrm{R}^{2}}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}\)

= \(\sqrt{I^{2} R^{2}+\left(I X_{L}-I X_{C}\right)^{2}}\)

= \(\sqrt{I^{2}\left[R^{2}+\left(X_{L}-X_{C}\right)^{2}\right]}=I \sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

As the frequency of AC source is increased, the capacitive reactance of the capacitor decreases and .inductive reactance of the .inductnr .increases . The values of X_{L} and X_{C} will be same at resonance frequency f_{r}.

X_{L} = X_{C}

2πfrL = \(\frac{1}{2 \pi \mathrm{f}_{\mathrm{r}} \mathrm{C}}\)

4π^{2} f_{r}^{2} LC =1

f_{r}^{2} = \(\frac{1}{4 \pi^{2} \mathrm{LC}}\), f_{r} = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

The current in RLC circuit will be maximum at resonance the variation of current with frequency is as shown below:

Question 6.

Explain the low pass filter with its frequency response.

Answer:

The low pass filter passes all the low frequencies below a cutoff frequency and stops or attenuates the higher frequencies.

The input is applied across the RC circuit and the output is taken across the capacitor.

At low frequencies, the capacitive reactance X_{C} = \(\frac{1}{2 \pi f \mathrm{c}}\) is high and hence entire input voltage appears across the output. At high frequencies, the capacitor offers very low reactance and acts as a short circuit and hence output becomes Zero. The cut off frequency f_{C} = \(\frac{1}{2 \pi R c}\)

Since the output voltage lags the input voltage, a low pass filter is also called a lag network.

Question 7.

Explain the high pass filter.

Answer:

A high pass filter passes all high frequencies above cut-off and rejects the lower frequencies.

At low frequencies, capacitor acts as open circuit (∴ X_{C} = \(\frac{1}{2 \pi \mathrm{RC}}\) = HIGH) At high frequencies, the capacitor acts as a short circuit. Hence all the input appears the output resistor. The cut off frequency, f_{C} = \(\frac{1}{2 \pi \mathrm{RC}}\)

As the output voltage leads the input voltage, the high pass filter is also called a lead network.