Class 12 Physics
CBSE and UP Board
Alternating Current
Chapter-7 Exercise 7.14
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 7 Question 14
Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
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Inductance of the inductor, L = 0.5 Hz, Resistance of the resistor, R = 100 Ω
Potential of the supply voltages, V = 240 V,
Frequency of the supply,ν = 10 kHz = 104 Hz
Angular frequency, ω = 2πν= 2π x 104 rad/s
Ans (a).
Peak voltage, V0 = √2 V= 240 √2 V
Maximum Current, I0 = V0 /√ (R² + ω² L²)
=240 √2/√[(100)² + (2π x 104 )² x (0.50)² ]
=1.1 x 10⁻²A
Ans (b).
For phase difference φ,we have the relation :
tanφ = Lω/R =(2π x 104 x 0.5)/100 = 100π
=> φ = 89.82º =89.82π /180 rad
ωt = 89.82π /180
=> t = 89.82π /(180 x 2π x 104) = 25μs
It can be observed that Io is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.