AC Voltage Applied to a Resistor: AC Voltage and Equations

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Jasmine Grover

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When voltage changes its direction after every half cycle is known as alternating voltage. The current flows in the circuit at that time are known as alternating current. The alternating current(AC) follows the sine function which changes its polarity concerning time. Most of the electrical devices are operating on the ac voltage.

Resistance is a substance that produces obstructions in the flow of electric current. It is usually denoted by R and its SI unit is Ohm. In DC voltage the resistance and current are related to each other using Ohm’s Law. In this article, we will understand how to apply Ohm’s law when the voltage applied is in the form of AC.


AC Voltage Applied to a Resistor

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Consider an alternating current source that produces potential difference across both terminals and responsible for the flow of electric current. Its representation is given in the figures below:

Alternating Current and Electromagnetic Induction
Alternating Current and Electromagnetic Induction

Let the voltage be given by:

V = Vsin ωt ----(1)

Here V0 = maximum amplitude of voltage 

ω = angular frequency

On applying Kirchhoff’s law to find the value of current in the resistance, we get:

∑ V(t) = 0

Equation (1) hence becomes:

V0sin ωt = I. R

Or

I = V0sin ωt / R 

According to Ohm’s Law:

V = I. R Or V/R = I

Then,

V0/R = I0

Hence, we obtain the following relationship between current, resistance and voltage in an AC circuit:

I = I0sin ωt

The above equation shows that Ohm’s law is equally working on the AC and DC voltages. The values of AC and DC reach their maximum and minimum simultaneously. The current and voltage in the ac circuit with resistor follow sinusoidal wave function having in the same phase.

The video below explains this:

AC voltage applied to a resistor Detailed Video Explanation:

Also Read:


Average Value of Current in R – Circuit

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The average or mean value of an alternating current is the total current flown in one complete cycle in a given time dividing by the time required to complete one cycle.

In other words, the average alternating current value is the average of every instantaneous value of AC from origin to its peak value.

iavg = 0Ti (t) dt/ 0Tdt

iavg = 1/T 0T i (t) dt

iavg = 1/T 0/ 2 i0sin ωtdt +1/T  T/2T i0sin ωtdt

iavg = i0/T (- cos ωt/ω)0/2 + i0/T (-cos ωt/ω)0/2T

iavg = i0/Tω (cosπ – cos0 – cosπ + cos2π)

Hence

iavg = 0

It is clear from the above equation that the average value of alternating current in one complete cycle is zero because it is positive in the first half cycle and negative in a second-half cycle with equal magnitude.


Root Mean Square (RMS) Value of Current in R – Circuit

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Root mean square or RMS value is defined as the amount of direct current passing through the resistance in the circuit for a given time that will produce the same amount of heat as the alternating current does in the same resistance at the same time. Mathematical derivations of Irms is given below:

Root Mean Square (RMS) Value of Current in R – Circuit
Root Mean Square (RMS) Value of Current in R – Circuit

Average value of power in AC circuit having a resistor

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p = i2. R = (i0)2.R sin2 ωt

here i0 and R are constants

using the trigonometric equation we have

sin 2ωt = ½ (1 – cos ωt)

since cos 2ωt = 0 then we can write

sin 2 ωt = ½

p = ½ i0². R

The presentation below highlights these basic concepts:

Also Read:


Sample Questions

Ques. Derive the relationship between peak and RMS value of current. (CBSE Delhi 2009)(5 Marks)

Ans: RMS value of current is defined as the amount of direct current passing through the resistance in the circuit for a given time that will produce the same amount of heat as the alternating current does in the same resistance at the same time. Mathematical derivations of Irms is given below:

irms derivation

Ques. Distinguish between ‘mean value’ and ‘RMS value’ of current. The instantaneous current from an AC source is I = 5sin(314t) A. What is the average and RMS value of current? (CBSE Delhi 2007)(3 marks)

Ans. RMS value of current is defined as the amount of direct current passing through the resistance in the circuit for a given time that will produce the same amount of heat as the alternating current does in the same resistance at the same time

On the other hand, the mean alternating current value is the average of every instantaneous value of AC from origin to its peak value.

As given in the question, Io = 5 A

RMS current = 0.707I= 3.53 A

Average current over the entire cycle is zero 

Ques. An alternating voltage given by V = 140 sin314t is connected across a pure resistor of 50 ohm. Find
the frequency of the source.
the rms current through the resistor. (CBSE AI 2012)(2 marks)

Ans: (i) From the instantaneous current equation, 2πν = 314 

Hence, ν = 50 Hz

(ii) RMS current = 0.707I= 0.707 x (140/50) = 2 A

Ques. Define ‘RMS value’ of current. Give the relation between peak current and RMS value of current (CBSE AI 2010 C)(1 mark)

Ans: RMS value of current is defined as the amount of direct current passing through the resistance in the circuit for a given time that will produce the same amount of heat as the alternating current does in the same resistance at the same time. The relation between rms and peak value of current is:

RMS current = 0.707I0

Ques. The peak value of emf in ac is E. Write its (i) rms (ii) average value over a complete cycle. (Foreign 2011)(1 Mark)

Ans: (i) Erms = 0.707E0 = 0.707E

(ii) Average value over a complete cycle is zero because during half the cycle the EMF goes up and during the other half the EMF goes down. As a result, the average value of emf is zero.

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CBSE CLASS XII Related Questions

  • 1.
    Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).


      • 2.
        The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

          • \( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
          • \( \frac{q}{\pi \epsilon_0 l^2} \) pointing along AM
          • \( \frac{q}{2 \pi \epsilon_0 l^2} \) pointing along AM
          • Zero

        • 3.
          What is displacement current (\( i_d \))? Considering the case of charging of a capacitor, show that \( i_d = \varepsilon_0 \frac{d\Phi_E}{dt} \). What is the value of \( i_d \) for a conductor across which a constant voltage is applied?


            • 4.
              If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.


                • 5.
                  Write any two features of nuclear forces.


                    • 6.
                      Two parallel plate capacitors X and Y are connected in series to a 6 V battery. They have the same plate area and same plate separation but capacitor X has air between its plates, whereas capacitor Y contains a material of dielectric constant 4. Calculate the capacitances of X and Y, if the equivalent capacitance of the combination of X and Y is \( 4 \, \mu\text{F} \). Calculate the potential difference across the plates of X and Y.

                        CBSE CLASS XII Previous Year Papers

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