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NCERT Class XII Chapter
Alternating Current
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Question : 15 of 26
Marks: +1, -0
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, XCX_C = 12πfC\frac{1}{2\pi f C}
XCX_C = 12×π×60×100×106\frac{1}{2 \times \pi \times 60 \times 100 \times 10^{-6}} = 26.54 Ω
Impedence, Z = R2+XC2\sqrt{R^2 + X_C^2} = (40)2+(26.54)2\sqrt{(40)^2 + (26.54)^2} = 48 Ω
(a) Virtual current in the circuit, IvI_v = EvZ\frac{E_v}{Z} = 11048\frac{110}{48} = 2.29 A
Maximum current I0I_0 = Iv2I_v \sqrt{2} = 3.24 A
(b) tan ϕ = VCVR\frac{V_C}{V_R} = 1ωCR\frac{1}{\omega C R}
Phase lag ϕ = tan1(1ωCR)\tan^{-1}\left(\frac{1}{\omega C R}\right) = tan1(26.5440)\tan^{-1}\left(\frac{26.54}{40}\right)
ϕ = 33.56° = 0.186π radian
Time lag t = ϕ/w = 0.18π2π(60)\frac{0.18\pi}{2\pi(60)} = 1.5 ms
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