The direct current which would give the same | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The $r.m.s.$ (root mean square) value of a current is defined as the value of the steady direct current which,when flowing through a given resisto

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The $r.m.s.$ (root mean square) value of a current is defined as the value of the steady direct current which,when flowing through a given resistor for a given time,produces the same amount of heat as the alternating current does in the same resistor for the same time.For the given square wave current:$I(t) = 2 	ext{ A}$ for $0 $I(t) = -2 	ext{ A}$ for $0.01 The time period $T = 0.02 	ext{ s}$.The $r.m.s.$ current is given by the formula:$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2(t) dt}$$I_{rms} = \sqrt{\frac{1}{0.02} \left[ \int_{0}^{0.01} (2)^2 dt + \int_{0.01}^{0.02} (-2)^2 dt ight]}$$I_{rms} = \sqrt{\frac{1}{0.02} \left[ 4 	imes 0.01 + 4 	imes 0.01 ight]}$$I_{rms} = \sqrt{\frac{1}{0.02} \left[ 0.04 + 0.04 ight]}$$I_{rms} = \sqrt{\frac{0.08}{0.02}} = \sqrt{4} = 2 	ext{ A}$.Thus,the $r.m.s.$ current is $2 	ext{ A}$.

The direct current which would give the same heating effect in an equal constant resistance as the current shown in the figure,i.e.,the $r.m.s.$ current,is.....$A$

Similar Questions

In an a.c. circuit $I=100 \sin 200 \pi t$. The time required for the current to achieve its peak value will be

$A$ multimeter reads a voltage of a certain $A$.$C$. source as $ 100 \,V $. What is the peak value of voltage of the $A$.$C$. source (in $\,V$)?

The current in a circuit varies with time as $I = 2 \sqrt{t}$. The $rms$ value of the current for the interval $t = 2 \, s$ to $t = 4 \, s$ is:

The voltage of an $AC$ source varies with time according to the equation,$V = 100 \sin(100 \pi t) \cos(100 \pi t)$. Where $t$ is in seconds and $V$ is in volts. Then:

An $AC$ current is given by $I = I_0 + I_1 \sin \omega t$. Then its $rms$ value will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module