17.5 AC Through a Resistor

17.5 AC Through a Resistor

A resistor $R$ connected with an AC source is shown in Fig. 17.6 (a). The instantaneous voltage $V$ is given as:

The instantaneous current $I$ through the circuit is: $I = \dfrac{V}{R}$ Using Eq. 17.10; $I = \dfrac{V_{o} \sin \omega t}{R}$

where $I_{o} = \dfrac{V_{o}}{R}$ is the peak or maximum value of current.

(Fig. 17.6(a): Showing a resistor $R$ connected with an AC source)

It follows from Eqs. (17.10) and (17.11) that the instantaneous values of both voltage and current are sine functions which vary with time. Figure 17.6(b) shows that both voltage and current pass their minimum and maximum values at the same time and thus their instantaneous values are said to be in phase with each other. Also Fig. 17.6(c) shows $V_{R}$ and $I_{R}$ vectors for resistance. $V$ and $I$ are drawn parallel because there is no phase difference between them. The opposition to AC which the circuit presents is the resistance given by $R = \dfrac{V}{I}$

(Fig. 17.6(b): Graphical representation for purely resistive circuit showing $V$ and $I$ in phase.)

(Fig. 17.6(c): $V_{R}$ and $I_{R}$ vectors representation for resistance.)

The power dissipation is proportional to the square of the current and makes no difference whether the current is direct or alternating i.e., whether the sign associated with the current is positive or negative. However, the power dissipation produced by AC having maximum value $I_{o}$ is not the same as that produced by a direct current of maximum value $I_{o}$, because the alternating current is at this maximum value only for an instant during each half-cycle. Instantaneous power $P$ dissipated across a resistor in AC circuit is:

and average power $\langle P\rangle = \langle I^{2}R\rangle$ As $\langle I^{2}\rangle = \dfrac{I_{o}^{2}}{2} = I_{\mathrm{rms}}^{2}$ , so Average power $\langle P\rangle = \dfrac{I_{o}^{2}}{2}R = I_{\mathrm{rms}}^{2}R$ Also, it can be proved that; Average power $\langle P\rangle = I_{\mathrm{rms}} V_{\mathrm{rms}}$ It is to be noted that here $P$ is measured in watts $V$ in volts, $I$ in amperes and $R$ in ohms respectively, and the Eq. (17.12) for power holds good only when $V$ and $I$ are in phase.

For your information

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Example 17.3

A $1\,\mathrm{kW}$ heating element is connected to a $250\,\mathrm{V}$ AC supply voltage. Calculate the amount of current taken from the supply and the resistance of the element when it is hot.

Solution

Power $P = 1\,\mathrm{kW} = 1000\,\mathrm{W}$ and Applied voltage $V = 250\,\mathrm{V}$ We know that; $P = VI$ or $I = \dfrac{P}{V}$ $I = \dfrac{1000\,\mathrm{W}}{250\,\mathrm{V}} = 4\,\mathrm{A}$ $R = \dfrac{250\,\mathrm{V}}{4\,\mathrm{A}} = 62.5\,\Omega$ Also $V = I R$ which suggests that, $R = \dfrac{V}{I}$ $R = \dfrac{250\,\mathrm{V}}{4\,\mathrm{A}}$ $R = 62.5\,\Omega$