class 12
a circuit draws 330 w from a 110V 60 Hz AC- line.the power factor is 0.6 and the current lags the voltage.the capacitance of a series capacitor that will result in a power factor of 1 is equal to
1 31 micro farad
2 54 micro farad
3 151 micro farad
4201 micro farad
a circuit draws 330 w from a 110V 60 Hz AC- line.the power factor is 0.6 and the current lags the voltage.the capacitance of a series capacitor that will result in a power factor of 1 is equal to
1 31 micro farad
2 54 micro farad
3 151 micro farad
4201 micro farad
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Answer:
b
Explanation:
Resistance of circuit,
R = V2P=110×110330=1103Ω
1st case Power factor, cosϕ = 0.6
Since, current lags the voltage thus, the circuit contains resistance and inductance .
∴ cosϕ=RR2+X2L−−−−−−−√=0.6
⇒R2+X2L=(R0.6)2
⇒X2L=R2(0.6)2−R2
⇒X2L=R2×0.640.36
∴XL=0.8R0.6=4R3 ...... (i)
IInd case Now cosϕ=1 [Given]
Therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which
XL=XC
∴XC=4R3=43×1103=4409Ω [from Eq.(i)]
⇒12πfC=4409Ω
∴C=92×3.14×60×440 = 0.000054F = 54μF
Answer:
31 μ
F
54 μ
F
151 μ
F
201 μ
F
Answer :
B
Solution :
Resistance of circuit,
R = V2P=110×110330=1103Ω
1st case Power factor, cosϕ = 0.6
Since, current lags the voltage thus, the circuit contains resistance and inductance .
∴ cosϕ=RR2+X2L−−−−−−−√=0.6
⇒R2+X2L=(R0.6)2
⇒X2L=R2(0.6)2−R2
⇒X2L=R2×0.640.36
∴XL=0.8R0.6=4R3 ...... (i)
IInd case Now cosϕ=1 [Given]
Therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which
XL=XC
∴XC=4R3=43×1103=4409Ω [from Eq.(i)]
⇒12πfC=4409Ω
∴C=92×3.14×60×440 = 0.000054F = 54μ
F
Explanation:
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