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AC Voltage Applied To A Resistor | AC Circuit Containing Resistance Only

AC Voltage Applied To A Resistor | AC Circuit Containing Resistance Only :- Suppose an alternating current source is connected to a pure resistance R as shown in figure (a) below :

Let the alternating voltage be represented by the following equation :-

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(1)

If I is the value of the current in the circuit at any time t , then the potential difference across the resistance will be IR . Now since there are no other components in the circuit, the potential difference across the resistance at any time must be equal to the applied emf, i.e.

$\displaystyle IR={{E}_{0}}sin\text{ }\omega t$

$\displaystyle \Rightarrow I=\frac{{{E}_{0}}}{R}sin\text{ }\omega t={{I}_{0}}sin\text{ }\omega t$ …..(2)

Where $\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{R}$ is the peak value of the current.

Equation (2) is the expression for alternating current in the circuit. By comparing $\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{R}$ with Ohm’s law, we find that in an AC circuit the opposition in the path of current is R, which is the same resistance R that opposes the current in a DC circuit.

Thus, the behavior of resistance is the same in both AC and DC circuits.

Comparing equation (1) and equation (2), we find that E and I are in the same phase. Therefore, in an AC circuit containing only resistance (R), the alternating voltage (E) and the alternating current (I) are in the same phase, as shown in the above figure (c).

Phasor Diagram

(AC Voltage Applied To A Resistor | AC Circuit Containing Resistance Only)

Since E and I are in the same phase, therefore in figure (b) the voltage phasor E₀ and the current phasor I₀ are shown rotating together in the same direction. That is, across the resistor R, the phase difference between the alternating voltage (E) and the alternating current (I) is zero (ωt – ωt = 0).

The projections of the voltage phasor E₀ and the current phasor I₀ on the y-axis represent the instantaneous values of voltage and current, respectively, that is,

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ और $\displaystyle I={{I}_{0}}sin\text{ }\omega t$

Conductance

Conductance is the reciprocal of resistance. Conductance measures how easily current can flow through a material.

चालकत्व $\displaystyle K=\frac{1}{R}$

The unit of conductance is ohm^-1 (Ω^-1).

Average Power

Instantaneous power of a purely resistive circuit :

$\displaystyle P=VI=\left( {{E}_{0}}sin\text{ }\omega t \right)\times \left( {{I}_{0}}sin\text{ }\omega t \right)$

$\displaystyle P=\left( {{I}_{0}}Rsin\text{ }\omega t \right)\times \left( {{I}_{0}}sin\text{ }\omega t \right)$

$\displaystyle P=I_{0}^{2}Rsi{{n}^{2}}\omega t$

Now, over one complete cycle, the value of sin²ωt is 1/2; therefore, the average power of the circuit over one complete cycle is :

$\displaystyle {{P}_{avg.}}=\langle I_{0}^{2}Rsi{{n}^{2}}\omega t\rangle =I_{0}^{2}R\langle si{{n}^{2}}\omega t\rangle =I_{0}^{2}R\times \frac{1}{2}$

$\displaystyle \Rightarrow {{P}_{avg.}}={{\left( \frac{{{I}_{0}}}{\sqrt{2}} \right)}^{\text{2}}}\times R$

Now we know that the root mean square (RMS) value of alternating current,

$\displaystyle {{I}_{rms}}=\frac{{{I}_{0}}}{\sqrt{2}}$

therefore,

$\displaystyle \therefore {{P}_{avg.}}=I_{rms}^{2}R$ …..(3)

Along with that,

$\displaystyle {{E}_{rms}}={{I}_{rms}}R$

therefore,

$\displaystyle {{P}_{avg.}}=I_{rms}^{2}R=\left( {{I}_{rms}}R \right)\times {{I}_{rms}}$

$\displaystyle \therefore {{P}_{avg.}}={{E}_{rms}}{{I}_{rms}}$ …..(4)

Equations (3) and (4) are the expressions for the average power in a purely resistive AC circuit.

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AC Voltage Applied To A Resistor | AC Circuit Containing Resistance Only

Phasor Diagram

(AC Voltage Applied To A Resistor | AC Circuit Containing Resistance Only)

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