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Natalka [10]
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
Physics
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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Here metal lead is having an oxidation state of +4 called as Pb^{4+} cation and sulphur non metal has oxidation state of -2 called as S^{2-}. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral PbS_2

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3 years ago
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