A current of 2, mA was passed through an unkn | vedclass

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

Question

(A) Given: Current $I = 2\, mA = 2 	imes 10^{-3}\, A$,Power $P = 4.4\, W$.Using the formula $P = I^2 R$,we find the resistance $R$:$R = \frac{P}{I^2}

Vedclass · Accepted Answer

$11 	imes 10^{-5}\, W$

Vedclass · Answer

(A) Given: Current $I = 2\, mA = 2 	imes 10^{-3}\, A$,Power $P = 4.4\, W$.Using the formula $P = I^2 R$,we find the resistance $R$:$R = \frac{P}{I^2} = \frac{4.4}{(2 	imes 10^{-3})^2} = \frac{4.4}{4 	imes 10^{-6}} = 1.1 	imes 10^6\, \Omega$.Now,when a voltage $V = 11\, V$ is applied across this resistor,the new power $P'$ is given by:$P' = \frac{V^2}{R} = \frac{11^2}{1.1 	imes 10^6} = \frac{121}{1.1 	imes 10^6} = 110 	imes 10^{-6}\, W = 11 	imes 10^{-5}\, W$.

$A$ current of $2\, mA$ was passed through an unknown resistor which dissipated a power of $4.4\, W$. The dissipated power when an ideal power supply of $11\, V$ is connected across it is:

Similar Questions

Assertion : When current through a bulb decreases by $0.5\%$,the glow of bulb decreases by $1\%$.
Reason : Glow (Power) is directly proportional to the square of current.

Define kilowatt hour and power.

$A$ coil develops heat of $800 \, cal/sec$ when $20 \, V$ is applied across its ends. The resistance of the coil is .............. $\Omega$ $(1 \, cal = 4.2 \, J)$.

$5.0 \, A$ current is set up in an external circuit by a $6.0 \, V$ storage battery for $6.0 \, \text{minutes}$. The chemical energy of the battery is reduced by:

An electric bulb is rated $220\, V$ and $100\, W$. Power consumed by it when operated on $110\, V$ is ............. $W$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module