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3 years ago

6

A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

battery supplies energy at a rate of 25 W, how large is the resistance

1 answer:

harina [27]3 years ago

7 0

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

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$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

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$V_{k} = 0.273\,L$

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