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Question: A battery of \[10V\] and internal resistance \[0.5\Omega \] is connected across a variable resistanc...

A battery of 10V10V and internal resistance 0.5Ω0.5\Omega is connected across a variable resistance R. the value of R for which the power delivered is maximum is equal to
(A) 0.25Ω0.25\Omega
(B) 0.5Ω0.5\Omega
(C) 1Ω1\Omega
(D) 2Ω2\Omega

Explanation

Solution

Hint We know that according to the maximum power theorem the cell will deliver maximum power when the external resistance must be equal to the internal resistance of the cell.

Complete step by step answer
It is given in the question that a battery of 10V10V and internal resistance 0.5Ω0.5\Omega is connected across a variable resistance R then we have to find the value of R for which the power delivered will be maximum.
We know that the power of a cell is given by P=I2RP = {I^2}R , it means P=V2(r+R)2RP = \dfrac{{{V^2}}}{{{{\left( {r + R} \right)}^2}}}R .
We know that according to the maximum power theorem the maximum external power from a source with a finite internal resistance, the load resistance must be equal to the source resistance. In our case, if we want to deliver maximum output then in such case the value of R must be equal to the internal resistance of the battery which is 0.5Ω0.5\Omega .
Therefore, when the value of R is 0.5Ω0.5\Omega the output power of the cell will be maximum.

Thus, option B is correct.

Note
One can do the mistake that they may assume that the maximum power of the cell is delivered when the current is maximum in the circuit and for maximum current, the resistance must be minimum and tick option A as their response but it is noted that according to maximum power theorem cell will deliver maximum power when the value of R is 0.5Ω0.5\Omega .