Test Index

NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 15
Total: 26
A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Solution:  

Capacitive reactance, XC =
1
2πfC
XC =
1
2×π×60×100×10−6
 = 26.54 Ω
Impedence, Z = √R2+XC2 = √(40)2+(26.54)2 = 48 Ω
(a) Virtual current in the circuit, Iv =
Ev
Z
 =
110
48
 = 2.29 A
Maximum current I0 = Iv√2 = 3.24 A

(b) tan ϕ =
VC
VR
 =
1
ωCR
Phase lag ϕ = tan−1(
1
ωCR
) = tan−1(
26.54
40
)
ϕ = 33.56° = 0.186π radian
Time lag t = ϕ/w =
0.18π
2π(60)
 = 1.5 ms
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