NCERT Physics Solution : Alternating Current

Q.1. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle ?

Sol. (a) rms value of current , Iv = Ev/R

Iv = 220/100 = 2.2 A

(b)Power consumed , P = Iv2 R
P = (2.2)2 x 100 = 484 W

Q.2. (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the
peak current ?

Sol. (a) rms voltage , Ev = Eo/√2 = 300/1.414 = 212.1 V

(b) peak current , Io = √2 Iv = 1.414 x 10 = 14.14 A

Q.3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Sol. Given , L = 0.044 H , Ev = 220 V , frequency ν = 50 Hz.

Inductive Reactance , XL = ωL = 2πνL

rms value of current , Iv = Ev /XL

On putting given values we get ,

Iv = 15.9 A

Q.4. A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Sol. Given , Ev = 110 V , frequency ν = 60 Hz , C = 60 μF

Capacitive reactance , Xc = 1/ωC = 1/2πνC

rms value of current , Iv = Ev /Xc

On putting the given values , we get

Iv = 2.49 A

Q.5. In Exercises 7.3 and 7.4, what is the net power absorbed by each
circuit over a complete cycle. Explain your answer.

Sol. Power Pav = Ev Iv cosφ

For Q.3 (Inductive Circuit) current lags behind the emf by π/2

Hence ,Pav = Ev Iv cosφ = 220 x 15.9 x cos90° = 0

In Q.4 (pure capacitive circuit ) current leads emf by phase π/2

Hence ,Pav = Ev Iv cosφ = Ev Iv cos90° = 0

Next Page »