91Ó°ÊÓ

1. Resistance, $R=50.0\text{Ω}$
2. Capacitance,$C=20\text{μ¹ó}=20×{10}^{-6}\text{F}$
3. Amplitude of emf, ${∈}_m=12.0\text{V}$

An electric circuit composed of a resistor and capacitor connected in series, is known as a RC circuit. By equating the phasor lengths for ${\mathit{V}}_{\mathit{R}}$and ${\mathit{V}}_{\mathit{c}}$, we can find thedriving frequency. By using the formula for phase angle in RCcircuit, we can find the phase angle corresponding to the driving frequency. We can find the angular speed of rotation of phasor by substituting the driving frequency in the formula for angular speed. We can find the current amplitude I in a series RCcircuit by using its formula.

The angular frequency,

${\mathit{Ó\neg }}_{\mathit{d}}\mathbf{=}\mathbf{2}\mathit{Ï€}{\mathit{f}}_{\mathit{d}}$ ...(1)

The capacitive reactance of the capacitor,

${\mathit{X}}_{\mathit{C}}\mathbf{=}\frac{\mathbf{1}}{{\mathit{Ó\neg }}_{\mathit{d}}\mathit{C}}$ ...(2)

The current equation using Ohm’s law,

$\mathit{I}\mathbf{=}\frac{{\mathit{Î\mu }}_{\mathit{m}}}{\mathit{Z}}$ ...(3)

The impedance of theLCRcircuit for the driving frequency,

$\mathit{Z}\mathbf{=}\sqrt{{\mathit{R}}^{\mathbf{2}}\mathbf{+}{\left({\mathit{X}}_{\mathit{c}}\right)}^{\mathbf{2}}}$ …(4)

Phase angle for series circuit,

$\mathrm{ϕ}\mathbf{=}\mathbf{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{-}\frac{{\mathbf{χ}}_{\mathbf{c}}}{\mathbf{R}}\right)\mathbf{}$ …(5)

Here, R is the resistance of the resistor, Cis the capacitance of the capacitor, ${\mathit{Î\mu }}_{\mathit{m}}$is the emf in the circuit and ${\mathit{f}}_{\mathit{d}}$is the frequency.

$V_C$lags behind $V_R$ by the phase angle $\frac{\mathrm{Ï€}}{2}\text{rad}$