Talha's Physics Academy
Flow of AC Through a Pure Resistor
Video Lecture
Watch the complete video lecture below to visually understand how alternating current flows through a pure resistor and its phase characteristics.
Describe the Flow of AC Through a Resistor
Mathematical Derivation and Analysis
Consider a pure resistor of resistance $R$ connected to an alternating voltage source. At any instant $t$, the alternating potential difference $v$ across the resistor's terminals is expressed as:
$$v = v_0 \sin(\omega t) \quad \text{--- (i)}$$
where $v_0$ signifies the peak value of the alternating voltage, and $\omega$ is the angular frequency.
According to Ohm's law, the circuit current $i$ at any instant is given by:
$$i = \frac{v}{R} \quad \text{--- (ii)}$$
Substituting equation (i) into equation (ii), we obtain:
$$i = \frac{v_0}{R} \sin(\omega t)$$
Since the peak current $i_0 = \frac{v_0}{R}$, the equation can be written as:
$$i = i_0 \sin(\omega t)$$
Phase Relationship
The current achieves its maximum value when $\sin(\omega t) = 1$. From equations (i) and (ii), it is evident that both the applied voltage and the current reach their zero values, positive peaks, and negative peaks simultaneously. Therefore, the applied voltage and the current in a purely resistive circuit are completely in phase with each other (phase difference $\theta = 0^\circ$).
Phasor Diagram of a Resistive Circuit
The current $I_R$ flowing through a resistor $R$ and the voltage $V_R$ across the resistor share the same phase. This alignment is represented on a phasor diagram by drawing a current vector ($I_R$) that completely coincides with the voltage vector ($V_R$).
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