Question

A capacitor of capacitance 100 microfarad and a resistance of 100 ohm is connected in series with ac supply of 220 v 50hz. The current leads the voltage by

Answer 1

Answer:tan^−1 (1/π)

Explanation:

We know that in R-C circuit,

tan ° = XC/R. ( ° is 'fi')

From formula, XC = 1/(ω*C)

XC = 1/(2πf*C) = 1/(100π*0.0001) = 100/π

It is given that R=100Ω

tan ° = ({100/π}/100) = 1/π

Hence, ° = tan^−1 (1/π)

Answer 2

Given :

Capacitance (C) = 100 μF = 10⁻⁴ F

Resistance (R) = 100 ohm

To Find :

The angle by which the current leads the voltage

Solution :

If 'w' is the angular velocity of the coil

w = 2πn = 2×π×50 = 100π rad/sec

Capacitive Reactance ($X_c$) = $\frac{1}{Cw}$

                                           = $\frac{1}{10^-^4 * 100\pi }$

                                           = $\frac{100}{\pi }$

If the angle between the impedance and x-axis is Ф, thus,

tanФ = $\frac{X_c}{R}$

         = $\frac{1}{CwR}$

         = $\frac{1}{10^-^4 * \frac{100}{\pi } *100}$

         = π

∴   Ф  =  tan⁻¹π

Therefore, the current leads the voltage by tan⁻¹π.