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Capacitive Circuits Alternating Currents L-4
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Capacitive Circuits Alternating Current
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-010-0019.2.jpg)
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Capacitive Circuits Alternating Currents L-4
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Alternating Current/Voltages
- $I=\frac{V_m}{R} \sin \omega t$ - $=Im \sin \omega t$ - Currrent is in-phase with voltage.
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Capacitive Circuits Alternating Currents L-4
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Phaser (Resistive cct)
- $|\vec{OA}| = Vm$ - $|\vec{OB}| = Im$
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Capacitive Circuits Alternating Currents L-4
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Pure Inductive Circuit
- $< I > = 0$ 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 $\pi/2$.
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Capacitive Circuits Alternating Currents L-4
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Inductive Reactors
- $I=I_m \sin (\omega t - \frac{\pi}{2})$ - $\omega L=$ Inductive Reactors $=X_L$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-065-0538.8.jpg)
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Capacitive Circuits Alternating Currents L-4
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Example
- 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 $t=0$, $I=0$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-076-0676.8.jpg)
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Capacitive Circuits Alternating Currents L-4
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Example
- $V=25 \sin (\omega t)$ 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? - $V_m=25V$ - $X_L=\omega L=400 \times 10 = 4000 \Omega$ - $I_m=\frac{Vm}{\omega L}=\frac{25}{4000}=6.25 mA$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-095-0915.6.jpg)
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Capacitive Circuits Alternating Currents L-4
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Example
- $V(t)=-12.5 V$ - $\omega t=\frac{7 \pi}{6}$ and $\frac{11 \pi}{6}$ - The magnitude is increasing at $\omega t=\frac{7 \pi}{8}$ - $I(t)=6.25 \sin (\frac{7 \pi}{6}-\frac{\pi}{2})$
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Capacitive Circuits Alternating Currents L-4
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Purely Capacitive Circuit
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Charging current
- $\frac{dQ}{dt}=C\frac{dV}{dt}$ - $V(t)=\frac{Q(t)}{C}$ - $\therefore Q(t)=CV(t)=CV_m \sin \omega t$ - $I(t)=CV_m \omega \cos \omega t$ - $I_ m = V_ m.(C \omega)=V_m (C \omega) \sin (\omega t+ \pi/2)$ - $\frac{V_ m}{I_ m}=\frac{1}{C \omega}=\Chi_ e \equiv I_m \sin (\omega t + \pi/2)$
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Capacitive Circuits Alternating Currents L-4
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Purely Capacitive Circuit
- $\Chi_C=\frac{1}{\omega C}$ - $I_ m=\frac{V_ m}{\Chi_C}=V_ m . \omega_C$
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Capacitive Circuits Alternating Currents L-4
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Purely Capacitive Circuit
- $V(t)=V_m \sin \omega t$
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Capacitive Circuits Alternating Currents L-4
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Phasor Diagram
- Current leads the voltage by $\frac{\pi}{2}$. - I becomes maximum T/4 before its voltage does.
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Capacitive Circuits Alternating Currents L-4
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Phasor Diagram for Capacitive Circuit
- 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 $\frac{T}{2}$.
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Capacitive Circuits Alternating Currents L-4
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Phasor Diagram for Capacitive Circuit
- 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.
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Capacitive Circuits Alternating Currents L-4
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Example
- At the instant when V(t) = -12.5 V and is increasing in magnitude - What is the current? - $\Chi_C =\frac{1}{\omega C}=\frac{1}{400 \times 10^{-8}}=250\Omega$ - $I_m=\frac{V_m}{X_c}=\frac{25}{250}=0.1A$
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Capacitive Circuits Alternating Currents L-4
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Example
- What is the time Lag between voltage and current maximum? - $\omega = 400 = \frac{2 \pi}{T}$ - $\frac{T}{4}=\frac{\pi}{800}=3.9 mS$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-265-2653.2.jpg)
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Capacitive Circuits Alternating Currents L-4
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Example
- Volatage divider using capacitors - $\Chi_{C_1}=\frac{1}{\omega c_!}=250 \Omega$ - $\Chi_{C_2}=\frac{!}{\omega c_2}=125 \Omega$ - $\Chi_C=250+125=375 \Omega$ - $I_m=\frac{V_m}{X_c}=\frac{25}{375}A=67mA$
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Capacitive Circuits Alternating Currents L-4
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Example
- A $5 \mu F$ is connected to $V_{rms}=60$ V. Observed peak current to be 0.2 A. Find frequency of source. - $V_{rms}=I_{rms} \Chi_C$ - $\Chi_C=\frac{V_{rms}}{I_{rms}}=\frac{60}{0.3}=200 \Omega$ - $\frac{1}{\omega c}=200 \rightarrow \omega = 1000 rad/s = 2 \pi f$ - $f=\frac{1000}{2 \pi}\simeq 160 Hz.$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-300-3050.4.jpg)
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Capacitive Circuits Alternating Currents L-4
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Power in a Capacitive Circuit
- Instantaneous power - $P=I(t)V(t)$ - $=I_m V_m \sin \omega t . \cos \omega t$ - $=\frac{I_m V_m}{2} \sin (2 \omega t)$ - $\langle P \rangle = 0 =(C\frac{dV}{dt})V$ - $=\frac{d}{dt}(\frac{1}{2}CV^2)$ - $W=\frac{1}{2}CV_m^2 \sin^2 \omega t$
![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-07-chapter-04-capacitive-circuits-alternating-currents-l-4_10-i67tl_hqxgm-321-3262.8.jpg)
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Capacitive Circuits Alternating Currents L-4
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Power
- 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 $\frac{T}{2}$
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Capacitive Circuits Alternating Currents L-4
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