Question

If a direct current of value a ampere is superimposed on an alternating current $I = b\sin \omega t$ flowing through a wire, what is the effective value of the resulting current in the circuit?
A.${\left[ {{a^2} - \dfrac{1}{2}{b^2}} \right]^{\dfrac{1}{2}}}$
B.${\left[ {{a^2} + {b^2}} \right]^{\dfrac{1}{2}}}$
C.${\left[ {\dfrac{1}{2}{a^2} + {b^2}} \right]^{\dfrac{1}{2}}}$
D.${\left[ {{a^2} + \dfrac{1}{2}{b^2}} \right]^{\dfrac{1}{2}}}$

Answer 1

Start by finding the total current flowing in the circuit and for effective current use the rms value , which can be defined as the square root of the mean (average) value of the squared function of the instantaneous values, found out by the relation ${I_{rms}} = \sqrt {\dfrac{{\int\limits_0^T {{I^2}} dt}}{T}}$. Simplify the trigonometric functions and use their periodicity while integrating .

Complete answer:
Given, direct current =${I_{dc}} = a$ and alternating current =${I_{ac}} = b\sin \omega t$
It is given that there are two currents i.e. direct current(D.C.) and alternating current(A.C.) .So the total amount of current flowing will be the sum of both the currents. i.e.
$I = {I_{dc}} + {I_{ac}}$
Substituting the values in above equation , we get
$I = a + b\sin \omega t$
And we know , The effective value of a varying voltage or current is the rms value , found out by the relation
${I_{rms}} = \sqrt {\dfrac{{\int\limits_0^T {{I^2}} dt}}{T}}$
Substituting the values , we get
${I_{rms}} = \sqrt {\dfrac{{\int\limits_0^T {{{(a + b\sin \omega t)}^2}} dt}}{T}}$
Now , applying ${(a + b)^2} = {a^2} + 2ab + {b^2}$, we get
${I_{rms}} = \sqrt {\dfrac{{\int\limits_0^T {({a^2} + {b^2}{{\sin }^2}\omega t + 2ab\sin \omega t)} dt}}{T}}$
Simplifying by separating the terms as per property of integrals, we get
${I_{rms}} = \sqrt {\dfrac{1}{T}\left[ {\int\limits_0^T {{a^2}} dt + \int\limits_0^T {{b^2}{{\sin }^2}\omega t} dt + \int\limits_0^T {2ab\sin \omega t} dt} \right]}$
We know,
$\cos 2\theta = 1 - 2{\sin ^2}\theta \\\ \Rightarrow {\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2} \\\$
Substituting this , we get
${I_{rms}} = \sqrt {\dfrac{1}{T}\left[ {\int\limits_0^T {{a^2}} dt + \int\limits_0^T {{b^2}\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)} dt + \int\limits_0^T {2ab\sin \omega t} dt} \right]} \\\ {I_{rms}} = \sqrt {\dfrac{1}{T}\left[ {\int\limits_0^T {{a^2}} dt + \int\limits_0^T {\dfrac{{{b^2}}}{2}} dt - \int\limits_0^T {\dfrac{{{b^2}\cos 2\omega t}}{2}} dt + \int\limits_0^T {2ab\sin \omega t} dt} \right]} \\\$
We know that $\int\limits_0^T {{b^2}\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)} dt = 0$ and $\int\limits_0^T {2ab\sin \omega t} dt = 0$ as the average value of cos and sin is zero over a cycle.
Therefore, we have
${I_{rms}} = \dfrac{1}{T}\sqrt {\int\limits_0^T {{a^2}} dt + \int\limits_0^T {\dfrac{{{b^2}}}{2}} dt}$
On solving the integrals , we have
$\Rightarrow {I_{rms}} = \sqrt {\dfrac{1}{T}({a^2}T + \dfrac{{{b^2}}}{2}T)}$
On further simplification ,we get
${I_{rms}} = \sqrt {{a^2} + \dfrac{{{b^2}}}{2}}$
Therefore , The effective current is $\sqrt {{a^2} + \dfrac{{{b^2}}}{2}}$.

So , Option D is the correct answer.

Note:
Similar questions can be asked for average current or instantaneous current, which can be done by applying the relevant formula as per the standard definition. Attention must be given while integrating and dealing with periodic functions . Also students must remember the periodicity of certain functions such as sin ,cos and tan.