NCERT Class XII Chapter
Alternating Current
Questions With Solutions
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Question : 14
Total: 26
Obtain the answers (a) and (b) in Q. 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?
Solution:
At very high frequency, XL increases to infinitely large, hence circuit behaves as open circuit.
X L 10 3
(a) Current in the coil,I rms
Maximum current in the coil,
I 0 √ 2 I rms √ 2 ×
10 − 2
This current is extremely small. Thus, at high frequencies, the inductive reactance of an inductor is so large that it behaves as an open circuit.
(b) As tan ϕ =
Clearly, time lag =
10 − 6
In dc circuit (after steady state), v = 0 and as such XL = 0.
In this case, the inductor behaves like a pure resistor as it has no inductive reactance.
(a) Current in the coil,
Maximum current in the coil,
This current is extremely small. Thus, at high frequencies, the inductive reactance of an inductor is so large that it behaves as an open circuit.
(b) As tan ϕ =
Clearly, time lag =
In dc circuit (after steady state), v = 0 and as such XL = 0.
In this case, the inductor behaves like a pure resistor as it has no inductive reactance.
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