# AC Voltage Applied to a Resistor

Resistance is the opposition to the flow of current. AC is a current that changes its polarity. Hence, as we shall see a Resistor does offer resistance to AC. Here we will quantify this resistance and try to connect its value to the value of the resistance in a DC circuit. Let us see!

## Alternating Current

When a constant voltage source or battery is applied across a resistor current is developed in resister. This current has a unique direction and flows from the negative terminal of a battery to positive terminal. The magnitude of the current remains constant as well. If Direction of current through resistor changes periodically then current is called alternating current.

Square wave AC current

Sinusoidal AC Current

## Resistor in an AC circuit

To have a sinusoidal varying alternating current we need to have an AC voltage source because current is directly proportional to voltage. An AC generator or AC dynamo can be used as an AC voltage source.

Voltage V(t) is applied across resistance R. V(t) is sinusoidal voltage with peak Vm and time period T.

$$T\quad =\frac { 1 }{ f } =\frac { 2\pi }{ \omega }$$

Where f is frequency and ω is angular frequency. This kind of circuit is a purely resistive circuit. According to Kirchhoff’s law –

$$v(t)=Ri(t)\\ i(t)=\frac { v(t) }{ R } \\ i(t)=\frac { { V }_{ m }\sin { (ωt) } }{ R } \\ { i }_{ m }=\frac { { V }_{ m } }{ R } \\ i(t)={ i }_{ m }\sin { (ωt) }$$

Here voltage and current has same frequency and both are in same phase.

### Average Value of the Current

The average value of current can be found out by summing over the total change in the voltage and dividing it by the number of times we do the measurements. This can be done as:

$${ i }_{ avg }\quad =\quad \frac { \int _{ 0 }^{ T }{ i(t)dt } }{ \int _{ 0 }^{ T }{ dt } } \\ { i }_{ avg }\quad =\frac { 1 }{ T } \int _{ 0 }^{ T }{ i(t)dt } \\ { i }_{ avg }\quad =\quad \frac { 1 }{ T } \int _{ 0 }^{ T/2 }{ { i }_{ m }\sin { \omega t } dt } \quad -\frac { 1 }{ T } \int _{ T/2 }^{ T }{ { i }_{ m }\sin { \omega t } dt }$$

$${ i }_{ avg }\quad =\frac { { i }_{ m } }{ T } { \left[ \frac { \cos { \omega t } }{ \omega } \right] }_{ 0 }^{ T/2 }-\frac { { i }_{ m } }{ T } { \left[ \frac { \cos { \omega t } }{ \omega } \right] }_{ T/2 }^{ T }\\ { i }_{ avg }\quad =\frac { { i }_{ m } }{ T\omega } \left[ \cos { \pi } -\cos { 0 } -\cos { \pi } +\cos { 2\pi } \right]$$

Hence,  $${ i }_{ avg }\quad =\quad 0$$

Average value of a AC current over a cycle is zero because in 1st haft of time period current is positive and in 2nd half current is negative.

### Root Mean Square Value of Current

$${ { i }_{ rms } }^{ 2 }\quad =\quad \frac { \int _{ 0 }^{ T }{ { i }^{ 2 }(t)dt } }{ \int _{ 0 }^{ T }{ dt } } \\ { { i }_{ rms } }^{ 2 }\quad =\frac { 1 }{ T } \int _{ 0 }^{ T }{ { i }^{ 2 }(t)dt } \\ { { i }_{ rms } }^{ 2 }\quad =\quad \frac { 1 }{ T } \int _{ 0 }^{ T }{ { i }_{ m }^{ 2 }{ \sin { \omega t } }^{ 2 }dt } \quad \\ { { i }_{ rms } }^{ 2 }\quad =\quad \frac { { i }_{ m }^{ 2 } }{ 2T } \int _{ 0 }^{ T }{ (1-\cos { 2\omega t } )dt }$$

$${ { i }_{ rms } }^{ 2 }\quad =\frac { { i }_{ m }^{ 2 } }{ 2T } { \left[ t-\frac { \sin { 2\omega t } }{ 2\omega } \right] }_{ 0 }^{ T }\\ { { i }_{ rms } }^{ 2 }\quad =\frac { { i }_{ m }^{ 2 } }{ 2T } \left[ T-0-\frac { \sin { 2\omega T\quad – } \sin { 0 } }{ 2\omega } \right] \\ { { i }_{ rms } }^{ 2 }\quad =\frac { { i }_{ m }^{ 2 } }{ 2T } \left[ T-\frac { \sin { 4\pi } }{ 2\omega } \right] \\ { { i }_{ rms } }^{ 2 }\quad =\quad \frac { { i }_{ m }^{ 2 } }{ 2 }$$

Hence $${ { i }_{ rms } }\quad =\quad \frac { { i }_{ m } }{ \sqrt { 2 } } \\$$

## Solved Example for You

Q. A circuit has an ac voltage source of 100V and 50Hz frequency is and 1KΩ resister. Find peak and rms value of current. How much time current will take to reach its first negative peak?

Solution: Peak voltage  $${ V }_{ m\quad }=\quad 100volt$$

Peak current  $${ i }_{ m }=\frac { { V }_{ m } }{ R }$$

$${ i }_{ m }=\frac { 100 }{ 1000 } A\\ { i }_{ m }=0.1A$$

RMS value $${ { i }_{ rms } }\quad =\quad \frac { { i }_{ m } }{ \sqrt { 2 } } \\ { { i }_{ rms } }\quad =\quad \frac { 0.1 }{ \sqrt { 2 } } \quad =0.070A=70mA$$

Time taken to reach first negative peak:

$$t\quad =\quad \frac { 3T }{ 2 } =\quad \frac { 3 }{ 2\times 50 } =.03sec$$

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##### Alternating Current

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