A battery of e.m.f. 10, V and internal resist | vedclass

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(D) The power $P$ delivered to a variable resistance $R$ connected to a battery of $e.m.f.$ $E$ and internal resistance $r$ is given by the formula $P

Vedclass · Accepted Answer

$0.5$

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(D) The power $P$ delivered to a variable resistance $R$ connected to a battery of $e.m.f.$ $E$ and internal resistance $r$ is given by the formula $P = I^2 R$,where $I = \frac{E}{R+r}$.Substituting $I$,we get $P = \left( \frac{E}{R+r} \right)^2 R$.To find the value of $R$ for maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = 0$.This condition leads to the Maximum Power Transfer Theorem,which states that the power delivered to the load is maximum when the load resistance $R$ is equal to the internal resistance $r$ of the source.Given $r = 0.5\, \Omega$,the power delivered is maximum when $R = r = 0.5\, \Omega$.

$A$ battery of $e.m.f.$ $10\, V$ and internal resistance $0.5\, \Omega$ is connected across a variable resistance $R$. The value of $R$ for which the power delivered in it is maximum is given by ......... $\Omega$.

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