PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Notes : CBSE Class 12 Physics Chapter 7 Alternating Current - Physics Kund 1. What is A.C. (Alternating Current)? Its Advantages and Disadvantages 2. Alternating Voltage Applied to a Resistor 3. Alternating Voltage Applied to an Inductor 4. Alternating Voltage Applied to a Capacitor 5. Alternating Current: Peak Value and RMS Value of Alternating Current/Voltage 6. Reactance and Impedance 7. LCR Series Circuit (Phasor Diagram Only) 8. Resonance in LCR Series Circuit 9. Power in AC Circuits 10. Power Factor 11. Wattless Current 12. AC Generator 13. Transformer

ALTERNATING VOLTAGE APPLIED TO A CAPACITOR Alternating source connected to a capacitor of capacitance C. Such a circuit is known as purely capacitive circuit. The capacitor is periodically charged and discharged when alternating voltage is applied to it. The alternating voltage applied across the capacitor is given by   $V = V_0 \sin \omega t \quad 1$ Let q be the charge on the capacitor at any instant. Then, potential difference across the capacitor,   $V_C = \frac{q}{C}$ But   $V_C = V \quad \text{or} \quad \frac{q}{C} = V = V_0 \sin \omega t$ $\therefore q = V_0 C \sin \omega t$ Now,   $I = \frac{dq}{dt} = \frac{d}{dt}(V_0 C \sin \omega t)$ $I= V_0 C (\cos \omega t) \omega$ $I= \frac{V_0}{(1 / C \omega)} \cos \omega t$ Since   $\cos \omega t = \sin \left(\omega t + \frac{\pi}{2}\right)$ $I= \frac{V_0}{(1 / C \omega)} \sin \left(\omega t + \frac{\pi}{2}\right)$ $\therefore I = I_0 \sin \left(\omega t + \frac{\pi}{2}\right) \quad ...(2)...

Alternating Voltage Applied to an Inductor :  An alternating source is shown connected to an ideal inductor of inductance (L). Such a circuit is known as purely inductive circuit. The alternating voltage across the inductor is given by   $V = V_0 \sin \omega t \quad \text{…(i)}$ The induced e.m.f. across the inductor =  $- L \frac{dI}{dt}$ which opposes the growth of current in the circuit. As there is no potential drop across the circuit, so,   $V + \left(-L \frac{dI}{dt}\right) = 0$ $L \frac{dI}{dt} = V$ $\frac{dI}{dt} = \frac{V}{L}$ Using eqn. (i), we get,   $\frac{dI}{dt} = \frac{V_0}{L} \sin \omega t$ $dI = \frac{V_0}{L} \sin \omega t \, dt \quad \text{…(ii)}$ Integrating both sides, we get,   $\int dI = \int \frac{V_0}{L} \sin \omega t \, dt = \frac{V_0}{L} \int \sin \omega t \, dt$ or   $I = \frac{V_0}{L} \left(-\frac{\cos \omega t}{\omega}\right) = \frac{V_0}{L \omega} (-\cos \omega t)$ Since   $(-\cos \omega ...

Alternating Voltage Applied to a Resistor :  The applied alternating voltage is given by,   $V = V_0 \sin \omega t$ Let ( I ) be the current in the circuit at any instant ( t ). As per Ohm's Law, potential difference across the resistor,   $V = IR$ or   $I = \frac{V}{R}$ Using eqn. (i), we get,   $I = \frac{V_0 \sin \omega t}{R}$ or   $I = I_0 \sin \omega t$ where,   $I_0 = \frac{V_0}{R}$ $I_0$ is the peak value of alternating current. Phasor diagram for pure resistive circuit shows that the phase difference between current ( I ) and voltage ( V ) is zero. Instantaneous Power :  The instantaneous power dissipated in resistor as per Joule’s heating effect is given by   $P = I^2 R = (I_0^2 \sin^2 \omega t) R$ Average Power :  $P_{av} = \text{average of } I^2 R$ $P_{av}= \text{average of } I_0^2 R \sin^2 \omega t$ Since,   $\sin^2 \omega t = \tfrac{1}{2}(1 - \cos 2\omega t)$ Average of ( $\cos 2\...

Question: What do you understand by the term A.C.? Alternating Current (A.C.): An electric current whose magnitude changes with time and polarity reverses periodically is called alternating current (A.C.). The instantaneous value of alternating current (a.c.) is expressed as: $I = I_0 \sin \omega t \quad ...(1)$ where ($I_0$) is the maximum value or peak value or amplitude of a.c. and ( $\omega$) is the angular velocity, where: $\omega = \frac{2\pi}{T} = 2\pi v$ Here, (T) is the time period and ($\nu$) is called the frequency of A.C. The instantaneous value of voltage of an A.C. source is given by: $V = V_0 \sin \omega t \quad ...(2)$ where ( $V_0$) is the maximum value or peak value or amplitude of voltage and ($\omega$) is driving angular velocity. Note: 1. An alternating current and alternating voltage are preferably represented by small letters as ( $i = i_m \sin \omega t$) and ( $v = v_m \sin \omega t$) respectively, where ($ i_m$) and ( $v_m$) are the peak values of alternating c...