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A coil of inductive reactance 31Ω31 \, \Omega has a resistance of 8Ω8 \, \Omega. It is placed in series with a condenser of capacitive reactance 25Ω25 \, \Omega. The combination is connected to an a.c. source of 110110 V. The power factor of the circuit is:

A

0.56

B

0.64

C

0.8

D

0.33

Step-by-Step Solution

The power factor (cosϕ\cos \phi) of a series LCR circuit is defined as the ratio of resistance (RR) to impedance (ZZ), given by cosϕ=RZ\cos \phi = \frac{R}{Z} . First, calculate the impedance ZZ. The formula for impedance in a series LCR circuit is Z=R2+(XLXC)2Z = \sqrt{R^2 + (X_L - X_C)^2} . Given: Resistance R=8ΩR = 8 \, \Omega Inductive Reactance XL=31ΩX_L = 31 \, \Omega Capacitive Reactance XC=25ΩX_C = 25 \, \Omega Calculate the net reactance: X=XLXC=3125=6ΩX = X_L - X_C = 31 - 25 = 6 \, \Omega. Calculate the impedance: Z=82+62=64+36=100=10ΩZ = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \Omega. Calculate the power factor: cosϕ=RZ=810=0.8\cos \phi = \frac{R}{Z} = \frac{8}{10} = 0.8.

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