A direct current of 5 A is superposed on an alternating current I=10 ω t flowing through the wire. The effective value of the resulting current will be

Correct answer: B. 5 3 A Total carrent, 1=(5+10 ω t) aligned _ eff &=[ ∫_ 0 ^ T I^ 2 dt ∫_ 0 ^ T dt ]^ 1 / 2 &=[ 1 ~T ∫_ 0 ^ T (5+10 ω t)^ 2 dt]^ 1 / 2 &=[ 1 ~T ∫_ 0 ^ T (25+100 ω t+100 ^ 2 ω t)]^ 1 / 2 But, 1 ~T ∫_ 0 ^ T ω t dt=0 aligned and array l 1 ~T ∫_ 0 ^ T ^ 2 ω t dt=(1)/(2) So, I_ eff =[25+(1)/(2) ×100]^ 1 / 2 =5 3 ~A array

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Alternating Current

BITSAT 2015 · Physics · Alternating Current

A direct current of 5 A is superposed on an alternating current...

A direct current of

5 \mathrm{ A}

is superposed on an alternating current

I=10 \sin \omega t

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

Options

A $(15 / 2) \mathrm{A}$
B $5 \sqrt{3} \mathrm{ A}$
C $5 \sqrt{5} \mathrm{ A}$
D $15 \mathrm{ A}$

Correct answer

B. $5 \sqrt{3} \mathrm{ A}$

Step-by-step solution

Total carrent,

1=(5+10 \sin \omega \mathrm{t})

\begin{aligned} \RightarrowI_{\text{eff }} &=\left[\frac{\int_{0}^{\mathrm{T}} \mathrm{I}^{2} \mathrm{dt}}{\int_{0}^{\mathrm{T}} \mathrm{dt}}\right]^{1 / 2}\\ &=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}(5+10 \sin \omega \mathrm{t})^{2} \mathrm{dt}\right]^{1 / 2}\\ &=\left[\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}}\left(25+100 \sin \omega \mathrm{t}+100 \sin ^{2} \omega \mathrm{t}\right)\right]^{1 / 2}\\ \text{ But, } \quad \frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin \omega \mathrm{t} \mathrm{dt}=0 \end{aligned}

and

\begin{array}{l}\frac{1}{\mathrm{~T}} \int_{0}^{\mathrm{T}} \sin ^{2} \omega \mathrm{t} \mathrm{dt}=\frac{1}{2}\\ \text{ So, } \mathrm{I}_{\mathrm{eff}}=\left[25+\frac{1}{2} \times100\right]^{1 / 2}=5 \sqrt{3} \mathrm{~A} \end{array}

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