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. $\sqrt{a^2+b^2}$
• B. $\sqrt{a^2+\frac{b^2}{4}}$
• C. $\sqrt{a^2+\frac{b^2}{2}}$
• D. $\sqrt{\frac{a^2}{2}+b^2}$

Answer 1

The correct option is C $\sqrt{a^2+\frac{b^2}{2}}$

The total current at any instant in the circuit will be,

$I=I_{DC}+I_{AC}=a+b\sin \omega t$

$\therefore I_{eff}=I_{rms}=\sqrt{\frac{{\int }_0^T{I}^{2}dt}{T}}$

$=\sqrt{\frac{{\int }_0^T(a+b\sin \omega t{)}^{2}dt}{T}}$

$=\sqrt{\frac{{\int }_0^T(a^2+2ab\sin \omega t+b^2{\sin }^2\omega t)dt}{T}}$

$=\sqrt{\frac{{\int }_0^T{a}^{2}dt+{\int }_0^{T}2ab\sin \omega tdt+{\int }_0^T{b}^2{\sin }^2\omega tdt}{T}}$

As,
$\frac{1}{T}{\int }_0^T\sin \omega tdt=0\text{and}\frac{1}{T}{\int }_0^T{\sin }^2\omega tdt=\frac{1}{2}$

$\therefore I_{eff}=\sqrt{\frac{{\int }_0^T{a}^{2}dt}{T}+\frac{{\int }_0^T{b}^{2}dt}{2T}}=\sqrt{a^2+\frac{b^2}{2}}$

Hence, option $\left(C\right)$ is correct answer.