Alternating Current Formula

The formulas for alternating current (AC) primarily describe how voltage and current fluctuate over time. AC analysis focuses on key aspects such as instantaneous values, the relationship between peak and RMS (Root Mean Square) values, and the calculation of power. These fundamental formulas are essential for understanding the behavior of AC circuits, forming the basis for electrical engineering in areas like power generation, transmission, and consumption. They help engineers and technicians model and analyze the performance of electrical systems in both residential and industrial applications.

1.0Equation for Alternating Current

$I=I_0\mathrm{Sin}\omega t$

$I=I_0\mathrm{Cos}\omega t$

2.0Average Value of A.C

$I_{\text{avg }}=<I>=\frac{{\int }_{t_1}^{t_2}Idt}{{\int }_{t_1}dt}=\frac{\text{ Area underI-t graph }}{\text{ Time interval }}$

1. For one complete Cycle=0
2. For Half Cycle$=\frac{2I_0}{\pi }$
3. For Half Wave Rectifier $=\frac{I_0}{\pi }$
4. For Full Wave Rectifier $=\frac{2I_0}{\pi }$

Physical Significance of Average Current:

$i_{av}\times \Delta t={\int }_0^{t}i.dt$

$\Rightarrow$ Charge flow by DC of value $i_{av}$=Charge flow by AC of value i

3.0Root Mean Square (RMS) Value of Current

$I_{rms}=\frac{I_0}{\sqrt{2}}$

$I_{rms}=0.707I_0$

Significance of RMS Value:

$i_{rms}^2\times R\times \Delta T={\int }_0^T{i}^{2}Rdt$

Heat produced by DC=Heat produced by AC

4.0AC Circuit Containing Resistor

Current and Voltage are in phase

$E=E_0\mathrm{Sin}\omega t$

$I=I_0\mathrm{Sin}\omega t$

$I_0=\frac{E_0}{R}$

$\frac{I_0}{\sqrt{2}}=\frac{E_0}{R\sqrt{2}}\Rightarrow I_{rms}=\frac{E_{rms}}{R}$

5.0AC Circuit Containing Inductor

Current always lag behind the emf by a phase angle of $\frac{\pi }{2}$

$E=E_0\mathrm{Sin}\omega t$

$I=I_0\mathrm{Sin}\left(\omega t-\frac{\pi }{2}\right)$

Inductive Reactance $\left(X_L\right)=\omega L=2\pi fL$

6.0AC Circuit Containing Capacitor

Current always lead the emf by a phase angle of $\frac{\pi }{2}$

$E=E_0\mathrm{Sin}\omega t$

$I=I_0\mathrm{Sin}\left(\omega t+\frac{\pi }{2}\right)$

$\left(X_C\right)=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

7.0AC Circuit containing R,L,C Key Formula

8.0AC Circuit Containing RL,RC,LC Circuit

9.0AC Circuit Containing Series LCR Circuit

Impedance(Z)

$=Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$

$I=\frac{E}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}\mathrm{Tan}\varphi =\frac{X_L-X_C}{R}$

Resonance:

1. $X_L=X_C$
2. $V_L=V_C$
3. $\varphi =0(V\text{ and I are in phase })$
4. $Z_{\text{min }}=R(\text{ Impedance Minimum })$
5. $I_{\max }=\frac{V}{R}(\text{ Current Maximum })$
6. Resonance Frequency $\left(f_r=\frac{1}{2\pi \sqrt{LC}}\right)$

Half -Power Frequency:

${\omega }_{1}and{\omega }_2$ are half power frequency

${\omega }_1={\omega }_0-\Delta \omega$

${\omega }_2={\omega }_0+\Delta \omega$

Bandwidth:

$=2\Delta \omega ={\omega }_2-{\omega }_1$

At ${\omega }_{1}and{\omega }_2\Rightarrow P=\frac{P_{\max }}{2}$ and $I=\frac{I_{\max }}{2}$

${\omega }_1{\omega }_2=\frac{1}{LC}\Rightarrow$${\omega }_0=\sqrt{{\omega }_1{\omega }_2}$

${\omega }_2-{\omega }_1=\frac{R}{L}\Rightarrow 2\Delta \omega =\frac{R}{L}$

Quality Factor:

$Q=\frac{{\omega }_0}{\text{ Bandwidth }}=\frac{{\omega }_0}{2\Delta \omega }$

$Q=\frac{{\omega }_{0}L}{R}=\frac{1}{{\omega }_{0}RC}$

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

10.0Power In AC

$P_{\text{avg }}=V_{rms}{I}_{rms}\mathrm{Cos}\varphi$

$\mathrm{Cos}\varphi =\frac{R}{Z}=\text{ Power Factor of ac circuit }$

• For Capacitive Circuit, $P_{\text{avg }}=0$
• For Inductive Circuit, $P_{\text{avg }}=0$
• RMS Power, $P_{rms}=V_{rms}{I}_{rms}$

11.0Wattless Current

• That component of current in ac-circuit which is not active, the component is $I\mathrm{Sin}\varphi$
• Wattfull Power =$P_{\text{avg }}=V_{rms}{I}_{rms}\mathrm{Cos}\varphi$
• Wattless Power =$P_{\text{avg }}=V_{rms}{I}_{rms}\mathrm{Sin}\varphi$

12.0Sample Questions On Alternating Current Formula

Q-1. Find out instantaneous current value for $I=I_0\mathrm{Sin}\omega tatt=\frac{T}{8}$

Solution:

$I=I_0\mathrm{Sin}\omega t$

$t=\frac{T}{8}\Rightarrow I=I_0\mathrm{Sin}\left(\frac{2\pi }{T}\right)\left(\frac{T}{8}\right)=I_0\mathrm{Sin}\frac{\pi }{4}\Rightarrow I=\frac{I_0}{\sqrt{2}}$

Q-2. Find out instantaneous voltage for $V=200\mathrm{Sin}(400\pi t)$ at $t=\frac{1}{800}\mathrm{Sec}$

Solution:

$V=200\mathrm{Sin}(400\pi t)$

at $t=\frac{1}{800}\mathrm{Sec}\Rightarrow V=200\mathrm{Sin}400\pi \times \frac{1}{800}\Rightarrow V=200\text{ Volt }$

Q-3.Find the time taken by current to reach $\frac{I_0}{\sqrt{2}}$, if the frequency of current is 50Hz.

Solution:

$I=0,\varphi =0^{\circ }$

$I=\frac{I_0}{\sqrt{2}},\varphi =\frac{\pi }{4}$

$\Delta t=\frac{T}{2\pi }\times \frac{\pi }{4}=\frac{T}{80}$

$\Delta t=\frac{1}{8f}=\frac{1}{8\times 50}=2.5\mathrm{ms}$

Q-4.In an AC main supply is given to be 220 V what would be the average emf during a positive half cycle?

Solution:

$V_{\text{avg }}=\frac{2V_0}{\pi }=\frac{2\sqrt{2}{V}_{rms}}{\pi }=\frac{2\sqrt{2}\times 220}{3.14}=198\mathrm{V}$

Q-5.A Capacitor of 50 pF is connected to an ac source of frequency 1 khz. Calculate its reactance.

Solution:

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi \times 10^3\times 50\times 10^{-12}}=\frac{10^7}{\pi }\Omega$