AC Voltage Applied to a Resistor

Most electricity supplied to homes and industries is alternating current (AC), not direct current (DC). This preference exists for three key engineering reasons:byjus

• Transformers work only on AC voltage, which can be stepped up for long-distance transmission and stepped down for safe household use
• Energy losses are reduced at high voltage during transmission (P_loss = I²R; smaller I means smaller loss)
• AC generators are simpler and more efficient to build at a large scale than DC sources

The current whose magnitude changes with time and direction reverses periodically is called alternating current (AC).

The voltage that varies sinusoidally with time, expressed as v = v_m sin⁡ ωt, is called alternating voltage.

The square root of the mean of the squares of the instantaneous current values over one complete cycle, equal to i_m\[\sqrt{2}​\], is called the root mean square current (or effective current).

An AC voltage source is connected in series with a pure resistor R.

Applied AC Voltage:

Where:

• v_m​ = peak (maximum) voltage (V)
• ω = angular frequency (rad/s)
• t = time (s)

Step 1:  Apply Kirchhoff's Voltage Law (KVL):

Since the only element is resistor R:

Step 2: Substitute v = v_m sin⁡ ωt:

Step 3: Solve for current i:

Step 4: Define peak current imi_mim​:

Step 5: Final current equation:

Instantaneous Power

Using the trigonometric identity sin⁡²θ = \[\frac{1-\cos2\theta}{2}\]:

Average Power

The average of cos ⁡2ωt over a complete cycle = 0. Therefore:

The DC power formula is P = I²R. For AC to produce the same heating effect as DC, we define an equivalent "effective" value called the RMS value.

Derivation of I_rms:

Average power: \[\bar{P}=\frac{1}{2}i_{m}^{2}R\]

For equivalence with DC: \[|\bar{P}=I_{rms}^{2}R\]

\[I_{rms}^2R=\frac{1}{2}i_m^2R\]

\[I_{rms}=\frac{i_m}{\sqrt{2}}\approx0.707i_m\]

Similarly: \[{V_{rms}=\frac{v_{m}}{\sqrt{2}}\approx0.707v_{m}}\]

RMS vs. Peak — Comparison Table

Power Using RMS Values

\[\bar{P}=I_{rms}^2R=\frac{V_{rms}^2}{R}=V_{rms}\cdot I_{rms}\]

Problem: A light bulb is rated at 100 W and a 220 V AC supply. Calculate: (a) Resistance of the bulb, (b) Peak voltage of the supply, (c) RMS current through the bulb.

Given:

• Power P = 100 W

• RMS Voltage V_rms = 220 V

Solution:

(a) Resistance:

(b) Peak voltage:

(c) RMS current:

Answer: R = 484 Ω, v_m​ ≈ 311 V, I_rms​ ≈ 0.45 A