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, $I_{\text{eff}}=[\frac{{\underset{0}{\overset{T}{\int }}{I}^{2}dt]}^{1/2}}{\underset{0}{\overset{T}{\int }}dt}={\left[\frac{1}{T}\underset{0}{\overset{T}{\int }}(a+b\sin \omega t{)}^{2}dt\right]}^{1/2}$
i.e., $I_{\text{eff }}={\left[\frac{1}{T}\underset{0}{\overset{T}{\int }}\left(a^2+2ab\sin \omega t+b^2{\sin }^2\omega t\right)dt\right]}^{1/2}$
but as $\frac{1}{T}\underset{0}{\overset{T}{\int }}\sin \omega tdt=0$ and $\frac{1}{T}\underset{0}{\overset{T}{\int }}{\sin }^2\omega tdt=\frac{1}{2}$
So, $I_{\text{eff }}={\left[a^2+\frac{1}{2}{b}^2\right]}^{1/2}$