Question

If a direct current of value $a$ ampere is superimposed on an alternative current $I=b \sin \omega t$ flowing through a wire, what is the effective value of the resulting current in the circuit?

• A. $\left[a^{2}-\frac{1}{2} b^{2}\right]^{1 / 2}$
• B. $\left[a^{2}+b^{2}\right]^{1 / 2}$
• C. $\left[\frac{a^{2}}{2}+b^{2}\right]^{1 / 2}$
• D. $\left[a^{2}+\frac{1}{2} b^{2}\right]^{1 / 2}$

Answer 1

Answer: (D)

As current at any instant in the circuit will be
$I=I_{ dc }+I_{ ac }=a+b \sin \omega t$
so, $I_{ \text{eff} }=\left[\frac{\left.\int\limits_{0}^{T} I^{2} d t\right]^{1 / 2}}{\int\limits_{0}^{T} d t}=\left[\frac{1}{T} \int\limits_{0}^{T}(a+b \sin \omega t)^{2} d t\right]^{1 / 2}\right.$
i.e., $I_{\text {eff }}=\left[\frac{1}{T} \int\limits_{0}^{T}\left(a^{2}+2 a b \sin \omega t+b^{2} \sin ^{2} \omega t\right) d t\right]^{1 / 2}$
but as $\frac{1}{T} \int\limits_{0}^{T} \sin \omega t d t=0$ and $\frac{1}{T} \int\limits_{0}^{T} \sin ^{2} \omega t d t=\frac{1}{2}$
So, $ I_{ \text{eff }}=\left[a^{2}+\frac{1}{2} b^{2}\right]^{1 / 2}$