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 ...

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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.

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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 , $p_{a v} = I_{r m s}^{2} R = {0.3}^{2} \times 100 = 0.9 W$

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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.