SATHEE: alternating-currents Question 9

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alternating-currents Question 9

Question: Q. 4. An alternating voltage given by $V = 140$ sin

$314 t$ is connected across a pure resistor of $50 Ω$ . Find :

(i) the frequency of the source.

(ii) the rms current through the resistor.

A [O.D. I, II, III 2012]

Show Answer

Solution:

Ans. (i)

$\begin{aligned} 2 π f & = ω 2 π f & = 314 rad f & = 50 Hz \end{aligned}$

$\Rightarrow 2 π f = 314 rad s^{- 1}$

(ii)

$\begin{aligned} I_{r m s} = \frac{V_{r m s}}{R} where, V_{r m s} = \frac{V_{0}}{\sqrt{2}} & I_{r m s} = \frac{140}{\sqrt{2} \times 50} = 1.98 A \approx 2 A \end{aligned}$

[CBSE Marking Scheme 2012]

Alternating Currents

alternating-currents Question 9

Short Answer Type Question-II

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Alternating Currents

alternating-currents Question 9

Short Answer Type Question-II

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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