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NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 17
Total: 26
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements of frequency. Source has emf 230 V and L = 5.0 H, C = 80 μF, R = 40 Ω.
Solution:  
Resonating angular frequency
ω =
1
√LC
 =
1
√5×80×10−6
 = 50 rad s−1
∴ Resonance of L and C in parallel can be calculated

1
X
 =
1
XL
−
1
XC
 =
1
ωL
−ωC
Impedence of R and X in parallel is given by
1
Z
 = √
1
R2
+
1
X2
At resonating frequency of series LCR, XL = XC
So,
1
X
 =
1
XL
−
1
XC
 = 0
Thus, impedances Z = R and will be maximum. Hence, in parallel resonant circuit, current is minimum at resonant frequency.
Current through circuit elements
IR =
EV
R
 =
230
4
 = 5.75 A
IC =
Ev
XL
 =
230
ωL
 =
230
50×5
 = 0.92 A
IC =
Ev
XL
 =
230
(1∕ωC)
 = 230 × 50 × 80 × 10−6 = 0.92 A
Since, IL and IC are opposite in phase, so net current,
Iv = IR+IL+IC
Iv = 5.75 + 0.92 √2 sin(ωt – π/2) + 0.92 √2 sin(ωt + π/2)
Iv = 5.75 – 0.92 √2 cos ωt + 0.92 √2 cosωt
Iv = 5.75 A
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