$I=\frac{V_m}{R} \sin \omega t$

$=Im \sin \omega t$

Currrent is in-phase with voltage.

$|\vec{OA}| = Vm$

$|\vec{OB}| = Im$

$< I > = 0$ Over cycle

$P_ {av} = \frac{I_ {m}^2R}{2}=I_{rms}^2R$

$I_{rms}=\frac{Im}{\sqrt{2}}$

$V_{rms}=\frac{Vm}{\sqrt{2}}$

$I(t) = \frac{Vm}{\omega L} \sin (\omega t = \frac{\pi}{2})$

Current lags by $\pi/2$.

$I=I_m \sin (\omega t - \frac{\pi}{2})$

$\omega L=$ Inductive Reactors $=X_L$

A resistive circuit in which rms value of current is 5A. What is the value of current $\frac{1}{400}$ second after it becomes zero. (frequency f = 50 hz)

$I_m=5 \sqrt{2}A$

$I(t)=I_m. \sin (100 \pi t)$

$=5 \sqrt{2}. \sin (100 \pi t)$

At $t=0$, $I=0$

$V=25 \sin (\omega t)$ in $\omega=400 rad/s$ in 10H inductor.

At the instant when the voltage is -12.5 V and its magnitude is increasing.

What is current?

$V_m=25V$

$X_L=\omega L=400 \times 10 = 4000 \Omega$

$I_m=\frac{Vm}{\omega L}=\frac{25}{4000}=6.25 mA$

$V(t)=-12.5 V$

$\omega t=\frac{7 \pi}{6}$ and $\frac{11 \pi}{6}$

The magnitude is increasing at $\omega t=\frac{7 \pi}{8}$

$I(t)=6.25 \sin (\frac{7 \pi}{6}-\frac{\pi}{2})$

Charging current

$\frac{dQ}{dt}=C\frac{dV}{dt}$

$V(t)=\frac{Q(t)}{C}$

$\therefore Q(t)=CV(t)=CV_m \sin \omega t$

$I(t)=CV_m \omega \cos \omega t$

$I_ m = V_ m.(C \omega)=V_m (C \omega) \sin (\omega t+ \pi/2)$

$\frac{V_ m}{I_ m}=\frac{1}{C \omega}=\Chi_ e \equiv I_m \sin (\omega t + \pi/2)$

$\Chi_C=\frac{1}{\omega C}$

$I_ m=\frac{V_ m}{\Chi_C}=V_ m . \omega_C$

Current leads the voltage by $\frac{\pi}{2}$.

I becomes maximum T/4 before its voltage does.

I > 0 charging current, Between T = 0 to t = $\frac{t}{4}$ charging.

At t = $\frac{T}{4}$ fully charged and there is no current from $\frac{T}{4}$ to $\frac{T}{2}$.

At t = $\frac{T}{2}$ the fully plates are fully discharged.

Charging from t = $\frac{T}{2}$ to $\frac{3T}{4}$.

Current reverses and plates are charged oppositely

Discharging from $\frac{3T}{4}$ to T.

At the instant when V(t) = -12.5 V and is increasing in magnitude

What is the current?

$\Chi_C =\frac{1}{\omega C}=\frac{1}{400 \times 10^{-8}}=250\Omega$

$I_m=\frac{V_m}{X_c}=\frac{25}{250}=0.1A$

What is the time Lag between voltage and current maximum?

$\omega = 400 = \frac{2 \pi}{T}$

$\frac{T}{4}=\frac{\pi}{800}=3.9 mS$

Volatage divider using capacitors

$\Chi_{C_1}=\frac{1}{\omega c_!}=250 \Omega$

$\Chi_{C_2}=\frac{!}{\omega c_2}=125 \Omega$

$\Chi_C=250+125=375 \Omega$

$I_m=\frac{V_m}{X_c}=\frac{25}{375}A=67mA$

A $5 \mu F$ is connected to $V_{rms}=60$ V. Observed peak current to be 0.2 A. Find frequency of source.

$V_{rms}=I_{rms} \Chi_C$

$\Chi_C=\frac{V_{rms}}{I_{rms}}=\frac{60}{0.3}=200 \Omega$

$\frac{1}{\omega c}=200 \rightarrow \omega = 1000 rad/s = 2 \pi f$

$f=\frac{1000}{2 \pi}\simeq 160 Hz.$

Instantaneous power

$P=I(t)V(t)$

$=I_m V_m \sin \omega t . \cos \omega t$

$=\frac{I_m V_m}{2} \sin (2 \omega t)$

$\langle P \rangle = 0 =(C\frac{dV}{dt})V$

$=\frac{d}{dt}(\frac{1}{2}CV^2)$

$W=\frac{1}{2}CV_m^2 \sin^2 \omega t$

P(t) = I(t) V(t)$=\frac{I_m V_m}{2}\sin(2 \omega t)$

(i) First quarter cycle t = 0, t = T/4

$(\omega t = 0 \neq \pi/2);\sin(2 \omega t) \geq 0$

I > 0, V > 0, P > 0 : Energy is observed

(ii) From t = $\frac{T}{4}$ to $\frac{T}{2}$