NCERT Questions for Class 12 Physics Chapter 7 – Alternating Current

Important Questions with Solutions of Class 12 Physics Chapter 7 – Alternating Current

1) Can the transformer step up a DC voltage?

Ans – The current remains constant in the primary coil when passing a DC voltage over the primary transformer. So, the magnetic flux connected to the secondary is unchanged & their voltage will be zero. Hence, a transformer cannot step up a DC voltage.

2) A coil of inductance 44 mH is linked with 220 V, 50 Hz AC supply. Calculate the RMS value of the circuit current.

Ans – Inductance of inductor, L = 44mH = 44 x 10^-3 H

Supply voltage, V = 220V, Frequency, 𝛎 = 50Hz

Angular frequency ω = 2π𝛎

Inductive reactance, XL = X_L = ωL = 2π𝛎L = 2π x 50 x 44 x 10^-3Ω

Then, the RMS value will be,

Hence, the RMS value of the circuit current is 15.92A.

3) Imagine when the initial capacitor charge in Exercise 7.7 is 6 mC. Calculate the total energy preserved in the circuit upfront. Also, predict the total energy in the end stage.

Ans – Capacitance of the capacitor, C = 30μF = 30 x 10^-6 F

Inductance of the inductor, L = 27mH = 27 x 10^-3 H

Charge on the capacitor, Q = 6mC = 6 x 10^-3 C

The total energy preserved in the capacitor is given by,

Ultimately, the total energy will be the same due to the shared energy between the capacitor & the inductor.

4) A hydroelectric power plant has a water pressure head with a height of 300m & an available water flow of 100 m³ s⁻¹ – When the turbine generator has 60% efficiency with the electric power present in the plant (g = 9.8ms^-2).

Ans – Height of water pressure head, h = 300m

The volume of water flow per second, v = 100m³/s

Efficiency of turbine generator, n = 60% = 0.6

Acceleration due to gravity, g = 9.8m/s²

Density of water, p = 10³ kg/m³

Electric power in the plant = η * hpgv

= 0.6 x 300 x 10³ x 9.8 x 100

= 176.4 x 10⁵ W

= 176.4 MW

5) A horizontal straight wire 10 m in length extends over east & west and falls at 5.0 m/s speed right towards the horizontal element of the earth’s magnetic field of strength. i.e, 0.30 x 10^-4wb/m²

(a) Calculate the instantaneous value of the induced EMF in the wire.

(b) Predict the EMF direction

(c) Figure the highest potential of the wire

Ans – length of the wire, l = 10m

v = 5.0 m/s

B_H = 0.30 x 10^-4 wb/m²

(a) Induced emf E = B_Hl𝜈

E = 0.3 x 10^-4 x 10 x 5

E = 1.5 x 10^-3 V

(b) Induced EMF begins from west to east.

(c) The potential will be higher in the eastern region.

6) (a) Define the reason for long-distance electric power transmission at high AC voltage.

(b) An AC generator has a coil of 50 turns with a 2.5m² area that rotates at an angular speed of 60 rad/s in B = 0.37 of uniform magnetic field between 2 fixed pole pieces. As we know, R = 500Ω

(i) Then, calculate the highest current value produced from the generator.

(ii) Predict the coil orientation when B has got high & zero magnetic flux.

Will the generator operate properly when the coils remain stationary & the pole pieces are moved at the same speed?

Ans – (a) The transmission of electric power over longer distances using high AC voltage allows a small current to flow through the transmission line to eliminate the overall power loss i.e. I²R.

(b) n = 50, A = 2.5m², w = 60 rad/s, B = 0.3T, R = 500Ω

(i) ε₀ = nABw

ε₀ = 50 x 2.5 x 0.3 x 60

ε₀ = 2250V

I₀ = ε₀/R = 2250/500 = 4.5A

(ii) There will be a massive magnetic flux when the coil is placed at a vertical position & remains zero if it’s in a horizontal position.

(iii) The operation of a generator depends on the relative motion between the coil & magnet.

7) Using the labeled diagram, derive the principle, construction and working of the transformer.

Ans –

Principle – A transformer modifies low AC voltage to a much higher AC voltage or vice-versa. Depending on the principle of mutual induction – i.e. If a changing current is created on the adjacent coil, the EMF will be induced in a coil.

Construction – With 2 sets of coils wrapped on a soft iron core, one is primarily linked to an AC source. Meanwhile, the other one is secondary merged with the load.

Working – The input voltage will change frequently if an alternating EMF is transmitted across the primary coil. Due to the magnetic flux, the primary coil is modified often. This reshuffling magnetic flux merged with the secondary coil can create induced EMF in the secondary coil.

Es = Ns (dΦs/dt)…(1)

Ep = Np (dΦp/dt)…(2)

When all the magnetic flux is created in the primary coil will get merged with the secondary one. i.e, Φs/Φp

Equating 1 & 2,

Es/Ep = Ns/Np

Es = Ep*(Ns/Np)

Where Ns/Np = K (transformation ratio)

k>1 is for a step-up transformer

k<1 is for a step-down transformer

When there is no loss of energy, then

EsIs = EpIp

Therefore, Es/Ep = Is/Ip

8) Consider the following problems:

(a) In any AC circuit, the given instantaneous voltage equals the algebraic sum of the instantaneous voltages over the series of circuit elements. Will it be applicable for RMS voltage?

(b) A capacitor in the primary layer of an induction coil. Define!

(c) An applied voltage signal has a superposition of a DC voltage & an AC voltage of maximum frequency. The circuit also has an inductor & a capacitor available in series. Display the visibility of the DC signal over C & the AC signal over L.

(d) A series-based choke coil with a lamp is linked to a DC circuit to make it shine brightly by inserting an iron core in the choke due to no change in the lamp’s brightness. Predict the relevant observations if the connection is to an AC line.

(e) Determining why a choke coil is required for fluorescent tubes with AC mains. Will it give any success if we use an ordinary resistor instead of over the choke coil?

Ans – (a) The statement is not true for RMS voltage.

Yes, in any AC circuit, the applied voltage equals the average value of the instantaneous voltages over the series circuit elements. Meanwhile, it’s a false statement for RMS voltage since voltages over multiple elements cannot be in phase.

(b) For capacitor charging, the maximum induced voltage is deployed. In a primary circuit, an induction coil is incorporated for the capacitor. Since, if the circuit gets broken, a high induced voltage can recharge the capacitor to eliminate further sparks.

(c) The DC signal hovers over capacitor C since DC signals have the impedance of an inductor (L). It can be applicable if the impedance of a capacitor (C) is extremely massive (almost ∞). So, a DC signal is displayed over C. Meanwhile, for an AC signal of higher frequency, the impedance of L is massive compared to C, which is extremely low. Hence, an AC signal of high frequency is displayed over L.

(d) When an iron core is placed in the choke coil (series with a lamp linked to the AC line), it will glow on the lower side. It’s due to the choke coil & the iron core that leverages the circuit impedance.

(e) Fluorescent tubes operating with AC mains require a choke coil in the circuit as it reduces the voltage over the tube without consuming more power. Meanwhile, a normal resistor won’t replace a choke coil to serve this process as the power gets drained during heat dissipation.

9) A small village with an 800kW supply demand of electric power at 220V is kept at a 15km distance from an electric plant yielding 440 V power. The 2-wire line resistance carrying power is 0.5Ω/km. The village receives power from the same grid by a 4000- 220 V step-down transformer from a nearby village sub-station.

(a) Calculate the line power loss regarding sof heat.

(b) Find out the desired quantity required for the plant supply when there is considerable power loss regarding leakage

(c) Define the plant’s step-up transformer

Ans – The desired electric power, P = 800kW = 800 x 10³W

Supply voltage, V = 220V

The voltage of the electric plant generating power, V’ = 440V

The total distance between the village and the power station, d = 15km

Resistance of the 2 wire lines with power = 0.5Ω/km

Total wire resistance, R = (15+15)*0.5 = 15Ω

The sub-station deploys 4000 to 220V step-down transformer

Input Voltage V1 = 4000V

Output Voltage V2 = 220V

The RMS value of the current will be,

l = P/V1 = 800 x 10³/4000 = 200A

(a) Line power loss = I²R = (200)² x 15 = 600 x 103 W = 600W

(b) Imagine that the power loss is acceptable when the current leakage occurs:

Total power supply from plant = 800kW+600 kW = 1400kW

(c) Voltage drop occurred in the power line, IR = 200 x 15 = 3000V

So, the total voltage sent from the plant is 3000+4000 = 7000V

Also, the generated power is 440 V. Hence, the step-up transformer rating at the power plant is 440 to 7000V.