Students can Download Chapter 7 Alternating Current Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 7 Alternating Current

### Plus Two Physics Alternating Current NCERT Text Book Questions and Answers

Question 1.

A 100Ω resistor is connected to a 220V, 50 Hz ac supply.

- What is the rms value of current in the circuit?
- What is the net power consumed over a full cycle?

Answer:

Given R = 100Ω, E_{ν} = 220V, ν = 50 Hz.

1. Since l_{ν} = \(\frac{E_{ν}}{12}\)

So l_{ν} = \(\frac{220}{100}\) = 2.2 A

2. P_{aν} = E_{ν} I_{ν} = 220 × 2.2

or P_{aν} = 484 W.

Question 2.

- The peak voltage of an ac supply is 300V. What is the rms voltage?
- The rms value of current in an ac circuit is 10A. What is the peak current?

Answer:

1. Given E_{0} = 300 V, E_{aν} = ?

Since E_{ν} = \(\frac{E_{0}}{\sqrt{2}}\) = 0.707 × 300

or E_{ν} = 212.1V.

2. Given I_{ν} = 10A, I_{0}=?

Since l_{0} = \(\sqrt{2}\) E_{ν} = 1.414 × 10

or I = 14.14 A.

Question 3.

A 44 mH inductor is connected to 220V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

Given L = 44 mH = 44 × 10^{-3}H

E_{ν} = 220V, ν = 50Hz, I_{ν} = ?

Since I_{ν} = \(\frac{E_{ν}}{x_{L}}=\frac{220}{\omega L}\)

or Iν = 15.9A.

Question 4.

A 60µF capacitor is connected to a 11OV, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

Given C = 60µF = 60 × 10^{-6}F, E_{ν} = 110V, ν = 60 Hz.

Since I_{ν} = \(\frac{E_{ν}}{x_{c}}\)

∴ I_{ν} = ωCE_{ν}

= 2pνCE_{ν} = 2 × 3.142 × 60 × 60 × 10^{-6} × 110

= 2.49A or

I_{ν} = 2.49V.

Question 5.

In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer:

In both the cases the net power consumed is zero because in both the cases.

Net power consumed P = E_{ν} l_{ν}cosΦ

and Φ =90°

∴ P = 0 (in each case).

Question 6.

Obtain the resonant frequency of a series LCR circuit with L=2.0H, C=32µV and R = 10?. What is the Q-value of this circuit?

Answer:

Given L = 2.0H, C = 32µF = 32 × 10^{-6}F

R = 10Ω, Q = ?, ω_{0} = ?

Resonant frequency

Question 7.

A charged 30µF capacitor is connected to a 27mH inductor. What is the angular frequency of free oscillations of the circuit?

Answer:

Given C = 30µF=30 × 10^{-6}F,L = 27mH = 27 × 10^{-3}H

ω_{0} = ?

or ω_{0} = 1.1 × 10^{-3}S^{-1}.

### Plus Two Physics Alternating Current One Mark Questions and Answers

Question 1.

Which type of transformer you use to operate the coffee maker at 220 V?

Answer:

Step down transformer.

Question 2.

In an A C. circuit, I_{rms} and I_{o} are related as.

Answer:

(d) I_{rms} = \(\frac{I_{0}}{\sqrt{2}}\)

Question 3.

A capacitor of capacitance C has reactance X. If capacitance and frequency become double then reactance will be

(a) 4X

(b) X/2

(c) X/4

(d) 2X

Answer:

(c) Explanation

Question 4.

Fill in the blanks

- Impedance: admittance
- ……………..: conductance

Answer:

Resistance

Question 5.

Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp?

Answer:

No power is developed across the inductor as heat.

Question 6.

In an a.c circuit with phase voltage V and current I, the power dissipated is

(a) V.l

(b) Depends on phase angle between V and I

(c) \(\frac{1}{2}\) × V.I

(d) \(\frac{1}{\sqrt{2}}\)

Answer:

(b) Depends on phase angle between V and I

### Plus Two Physics Alternating Current Two Mark Questions and Answers

Question 1.

Fill in the blanks.

Answer:

(i) Current lags by π/2

(ii) X_{c} = 1/cω

(iii) R

(iv) Phase difference is zero.

Question 2.

A.C. adaptor converts household ac into low voltage dc. A stepdown transformer is a essential part of ac adapter.

- What is the use of step down transformer?
- What is the principle of a transformer? Explain.

Answer:

- To decrease voltage
- It works on the principle of mutual induction.

### Plus Two Physics Alternating Current Three Mark Questions and Answers

Question 1.

Match the following

Answer:

Question 2.

The following figure is a part of a radio circuit.

- Identify the circuit.
- What happens to this circuit if X
_{L}= X_{C} - lf X
_{C }> X_{L}draw the phaser diagram.

Answer:

1. LCR circuit.

2. When X_{L} = X_{C}, the impedance of circuit becomes minimum and the current corresponding to that frequency is maximum.

3.

### Plus Two Physics Alternating Current Four Mark Questions and Answers

Question 1.

Figure below shows a bulb connected in an electrical circuit.

1. When the key is switched ON the bulb obtains maximum glow only after a shorter interval of time which property of the solenoid is responsible for the delay?

- Self-induction
- Mutual Induction
- Inductive reactance
- None of the above

2. If the flux linked with the solenoid changes from 0 to 1 weber in 2 sec. Find the induced emf in the solenoid.

3. If the 3v battery is replaced by an AC source with the key closed, what will be observation? Justify your answer.

Answer:

1. Self-induction.

2. \(\frac{d \phi}{d t}=\frac{1}{2}\) = 0.5V.

3. When AC is connected the brightness of bulb will be decreased. This is due to the back emf in the circuit.

Question 2.

A graph connecting the voltage generated by an a.c. source and time is shown.

- What is the maximum voltage generated by the source?
- Write the relation connecting voltage and time
- This a.c. source, when connected to a resistor, produces 40J of heat per second. Find the equivalent d.c. voltage which will produce the same heat in this resistor.

Answer:

- 200v.
- V = V
_{0}sin ωt - \(V_{\max }=\frac{V_{0}}{\sqrt{2}}=\frac{200}{\sqrt{2}}\) = 141.8v.

Question 3.

An inductor, capacitor, and resister are connected in series to an a.c. source V = V_{0}sin ωt.

- Draw a circuit diagram of L.C.R. series circuit with applied a.c. voltage.
- Find an expression for impedance of L.C.R. series circuit using phasor diagram.
- What is impedance of L.C.R. series circuit at resonance?

Answer:

1.

2.

For impedance of LCR circuit.

From the right angled triangle OAE,

Final voltage, V = \(\sqrt{v_{n}^{2}+\left(v_{L}-v_{c}\right)^{2}}\)

Where Z is called impedance of LCR circuit

3. Impedance, Z=R.

Question 4.

A voltage source is connected to an electrical component X as shown in figure.

1. Identify the device X.

2. Which of the following equations can represent the current through the circuit?

- i = i
_{m }sin(ωt + π/2) - i = i
_{m}sin(ωt – π/2) - i = i
_{m}sin ωt - i = i
_{m }sin(ωt + π/4)

3. Draw the phasor diagram for the circuit. (2)

Answer:

1. Resistor

2. i = i_{m} sin ωt

3.

Question 5.

A friend from abroad presents you a coffee maker when she visited you. Unfortunately, it was designed to operate at 110V line to obtain 960W power that it needs.

- Which type of transformer you use to operate the coffee maker at 220V? (1)
- Assuming the transformer you use as ideal, calculate the primary and secondary currents. (2)
- What is the resistance of the coffee maker? (1)

Answer:

1. Step down transformer.

2. Since the transformer is ideal

V_{p}I_{p} = V_{s} I_{s} = 960W, V_{p} = 220v, V_{s} = 110v

V_{p}I_{p} =960

3.

### Plus Two Physics Alternating Current Five Mark Questions and Answers

Question 1.

The voltage-current values obtained from a transformer constructed by a student is shown in the following table.

- Identify the transformer as step up or step down.
- How much power is wasted by the transformer?
- What are the possible energy losses in a transformer?
- If the input voltage is 48v and input current is 1 A, is it possible to light 240v, 100w bulb using the above transformer. Justify.

Answer:

1. Step down transformer.

2. Power loss = 200w -10w = 190w

3. The possible energy losses in a transformer:

- Eddy current loss
- Copper loss
- Hysteries loss
- Flux leakage

4. In this case input power = VI = 48 × 1 = 48W.

If transformer does not waste energy, input power =out put power.

Hence maximum output power 48W. But bulb requires 100w. Hence the bulb does not glow with this low input power.

Question 2.

The current through fluorescent lights are usually limited using an inductor.

- Obtain the relation i = i
_{m }sin(ωt – π/2)for an inductor across which an alternating emf v = v_{m}sin ωt is applied. (2) - Why it is better to use an inductor rather than a resistor to limit the current through the fluorescent lamp? (1)
- When 100 V DC source is connected across a coil a current of 1 A flows through it. When 100V, 50 Hz AC source is applied across the same coil only 0.5 A flows. Calculate the resistance and inductance of the coil. (2)

Answer:

1. AC voltage applied to an Inductor

Consider a circuit containing an inductor of inductance ‘L’ connected to an alternating voltage. Let the applied voltage be

V = V_{0} sinωt…………(1)

Due to the flow of alternating current through coil, an emf, \(\mathrm{L} \frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}\) is produced in the coil. This induced emf is equal and opposite to the applied emf (in the case of ideal inductor).

2. No power is developed across the inductor as heat.

3. Resistance of the coil R = \(\frac{v}{I}=\frac{100}{1}\) = 100Ω. Current through the coil when ac source is applied.

R^{2} + X^{2}_{L} = 200^{2}

X^{2}_{L} = 200^{2} – 100^{2}

X_{L} = 173.2Ω

L_{ω} = 173.2

Question 3.

An alternating voltage is connected to a box with some unknown circuital arrangement. The applied voltage and current through the circuit are measured as v = 80 sinωt volt and i = 1.6 sin(ωt + 45°) ampere.

- Does the current lead or lag the voltage?
- Is the circuit in the box largely capacitive or inductive?
- Is the circuit in the box at resonance?
- What is the average power delivered by the box?

Answer:

1. Leads.

2. Capacitive

3. No Hint: Current and voltage are not in the same phase.

4. P = V_{rms}l_{rms}CosΦ, V_{m} = 80v, i_{m} = 1.6A

p = 45.25W