An alternating emf of 100 V-50 Hz is applied across a 10 - F capacitor and 100 - resistor in series- Calculate the reactance of the capacitor- the current flowing and the average power supplied-

According to the given data , reactance of the capacitance, #160; X_C=1ω C=12π× 50× 10× 10^ 6=318.5 Ω 'Hence , impedence , Z=√(R^2+X_C^2)=√(100^2+318 ...

You visited us 68 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Resistance of P-N Junction

An alternatin...

Question

An alternating emf of $100 V - 50 H z$ is applied across a $10 μ F$ capacitor and $100 Ω$ resistor in series. Calculate the reactance of the capacitor, the current flowing and the average power supplied.

Open in App

Solution

According to the given data , reactance of the capacitance, $X_{C} = \frac{1}{ω C} = \frac{1}{2 π \times 50 \times 10 \times 10^{- 6}} = 318.5 Ω$ '
Hence , impedence , $Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{100^{2} + {318.5}^{2}} = 333.89 = 334 Ω$
so, current flowing $I_{r m s} = \frac{V}{Z} = \frac{100}{334} = 0.299 = 0.30 A$
average power , $p_{a v} = I_{r m s}^{2} R = {0.3}^{2} \times 100 = 0.9 W$

Suggest Corrections

Similar questions

Q. A circuit is set up by connecting inductance L=100mH resistor R=100

Ω

and a capacitor of reactance 200

Ω

in series An alternating emf of 150

\sqrt{2} V, 500 m^{- 1}

Hz is applied across this series combination. Calculate the power dissipated.

An alternating emf of $e = 200 sin 100 π t$ is applied across a capacitor of capacitance $2 μ F$ . The capacitive reactance and the peak current is

Q. A

100 Ω

resistance and a capacitor of

100 Ω

reactance are connected in series across a

220 V

source. When the capacitor is 50% charged, the peak value of the displacement current is

Q. A resistor and a capacitor are connected to an A.C. supply of

200 V, 50 H z

in series. The current in the circuit is

2 A

. If the power consumed in the circuit is

100 W

then the capacitive reactance in the circuit is

When $100 V$ dc is supplied across a solenoid, a current of $1.0 A$ flows in it. When $100 V$ ac is applied across the same coil, the current drops to $0.5 A$ . If the frequency of ac source is $50 H z$ , then the impedance and inductance of the solenoid are

Join BYJU'S Learning Program

An alternating emf of 100 V−50 Hz is applied across a 10 μF capacitor and 100 Ω resistor in series. Calculate the reactance of the capacitor, the current flowing and the average power supplied.

According to the given data , reactance of the capacitance, XC=1ωC=12π×50×10×10−6=318.5Ω'Hence , impedence ,Z=√R2+X2C=√1002+318.52=333.89=334Ωso, current flowing Irms=VZ=100334=0.299=0.30Aaverage power ,pav=I2rmsR=0.32×100=0.9W

An alternating emf of $100 V - 50 H z$ is applied across a $10 μ F$ capacitor and $100 Ω$ resistor in series. Calculate the reactance of the capacitor, the current flowing and the average power supplied.