Question

A battery of emf $10\,V$ and internal resistance of $0.5$ ohm is connected across a variable resistance $R$ . The value of $R$ for maximum power transfer is given by:
A) $0.5\,\Omega$
B) $1.00\,\Omega$
C) $2.0\,\Omega$
D) $0.25\,\Omega$

Answer 1

Use the formula of the power and substitute the formula of the internal resistance in it. From the obtained formula of the power, determine the value of the resistance that must cause the maximum power transfer. This can be obtained by the greater value of the numerator than the denominator.

Formula used:
(1) The formula of the current is given by
$i = \dfrac{E}{{R + r}}$
Where $i$ is the current of the circuit, $E$ is the emf of the circuit, $R$ is the resistance of the circuit and $r$ is the internal resistance of the circuit.
(2) The formula of the power is given by
$P = VI$
Where $P$ is the power in the circuit, $V$ is the potential difference across the circuit and $I$ is the current.

Complete step by step solution:
It is given that the
Emf of the battery, $e = 10\,V$
Internal resistance, $r = 0.5\,\Omega$
Using the formula of the power,
$\Rightarrow P = Vi$
Substituting the ohm’s law in the above formula, we get
$\Rightarrow P = {i^2}R$
Again substituting the formula of the internal resistance in the above obtained formula,
$\Rightarrow P = {\left( {\dfrac{E}{{R + r}}} \right)^2}R$
By squaring the right hand side of the equation we get
$\Rightarrow P = \dfrac{{{E^2}R}}{{{{\left( {R + r} \right)}^2}}}$
The power transfer will be maximum when the $R = r$ that is the variable resistance must be equal to the internal resistance.
$\Rightarrow R = r = 0.5\,\Omega$
Hence the power value will be maximum at the resistance of $0.5\,\Omega$.

Thus the option (A) is correct.

Note: The power transfer depends on the current it carries and the potential difference developed across the circuit. The current will be maximum when the resistance and the internal resistance are the same. Since the power depends on the current, power value will also be maximum.