Physics · Alternating Current · NEET
In an AC circuit the current and voltage do not always peak at the same time; they are separated by a phase angle phi. Power is voltage times current, but because they are out of step, only the part of the current that is in phase with the voltage does useful work. That in-phase part is I_rms x cos(phi). So average power P = V_rms x I_rms x cos(phi). When phi = 0 (pure resistor), cos(phi) = 1 and all the power is used. This factor is why NEET problems always need the phase angle.
No. In a pure inductor the current lags the voltage by 90 degrees, and in a pure capacitor it leads by 90 degrees. Either way phi = 90 degrees, so cos(phi) = cos 90 = 0, and average power P = 0. Energy flows into the inductor or capacitor for half the cycle and flows back out in the next half, so the net energy used over one full cycle is zero. This zero-power current is called wattless current. Only the resistor dissipates energy as heat.
Instantaneous power p = v x i changes every moment during the cycle and is not useful for practical work. What matters is the average power over one complete cycle, because that is what heats a device or shows on your electricity bill. Averaging the sin and cos terms over a full cycle gives P = V_rms x I_rms x cos(phi). For NEET, unless the question says 'at this instant', you should always compute average power.
They are always consistent. Since cos(phi) = R/Z and I_rms = V_rms/Z, substituting gives P = V_rms x (V_rms/Z) x (R/Z) = V_rms^2 R / Z^2 = I_rms^2 x R. So the average power equals I_rms^2 R, meaning all the real power is dissipated in the resistance only. In numericals, P = I_rms^2 R is often the fastest route once you know I_rms and R.
An inductor 20 mH, a capacitor 50 microF and a resistor 40 ohm are connected in series across a source of emf V = 10 sin 340t. The power loss in the AC circuit is:
An inductor 20 mH, a capacitor 100 microF and a resistor 50 ohm are connected in series across a source of emf V = 10 sin 314t. The power loss in the circuit is:
The maximum power is dissipated for an AC in a/an
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Average power P = V_rms x I_rms x cos(phi), where V_rms and I_rms are the root-mean-square voltage and current and phi is the phase angle between them. It can also be written as P = I_rms^2 x R, since the resistor is the only element that dissipates energy.
The watt (W). One watt equals one joule per second. In AC, V_rms x I_rms alone gives the apparent power in volt-amperes (VA), and multiplying by cos(phi) gives the real power in watts.
Because the phase angle is 90 degrees, giving cos 90 = 0. Energy is stored and returned each cycle with no net loss, so no heat is produced. This current is called wattless current.
No. Apparent power = V_rms x I_rms (in VA). Real or average power = V_rms x I_rms x cos(phi) (in W). They are equal only when cos(phi) = 1, that is, a purely resistive circuit.
Not directly. You must convert peak to rms first: V_rms = V_0/root2 and I_rms = I_0/root2. If you use peaks, P = (1/2) V_0 I_0 cos(phi), because the two factors of 1/root2 give an extra 1/2.
