A direct current of 6 A is superimposed on an alternating current I=10 ω t flowing through a wire. The effective value of the resulting current will be

Correct answer: C. 9.27 Given, I=6+10 ω t aligned I_ eft & =[ .∫_0^T I^2 d t]^ 1 / 2 ∫_0^T d t =[(1)/(T) ∫_0^T(6+10 )^2 d t]^ 1 / 2 . & =[(1)/(T) ∫_0^T(36+120 +100 ^2 ) d t]^ 1 / 2 aligned But as, (1)/(T) ∫_0^T ω t d t=0 and (1)/(T) ∫_0^T ^2 ω t= (1)/(2) I_ eff =[36+(1)/(2) ×100]^ 1 / 2 =9.27 Thus, I_ eff =9.27 A.

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Physics
Chapter: Waves and Sound

BITSAT 2022 · Physics · Waves and Sound

A direct current of 6 A is superimposed on an alternating current...

A direct current of

6 \mathrm{ A}

is superimposed on an alternating current

I=10 \sin \omega t

flowing through a wire. The effective value of the resulting current will be

Options

A $5 \sqrt{2}$
B $5 \sqrt{3}$
C $9.27$
D $8.37$

Correct answer

C. $9.27$

Step-by-step solution

Given,

I=6+10 \sin \omega t

\begin{aligned}I_{\mathrm{eft}} & =\left[\frac{\left.\int_0^T I^2 d t\right]^{1 / 2}}{\int_0^T d t}=\left[\frac{1}{T} \int_0^T(6+10 \sin \omegat)^2 d t\right]^{1 / 2}\right. \\ & =\left[\frac{1}{T} \int_0^T\left(36+120 \sin \omegat+100 \sin ^2 \omegat\right) d t\right]^{1 / 2} \end{aligned}

But as, \frac{1}{T} \int_0^T \sin \omega

t d t=0 and

\frac{1}{T} \int_0^T \sin ^2 \omega

\frac{1}{2} \Rightarrow \quad

I_{

\mathrm{eff} }=\left[36+\frac{1}{2} \times100\right]^{1 / 2}=9.27 Thus,

\quad I_{

\text{eff } }=9.27 \mathrm{ A}.

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