# Question 200c2

Feb 22, 2017

${I}_{\max} = 12 A$

#### Explanation:

$P = V I - R {I}^{2}$ and

$0 \le I \le 15$

The maximum value for $P$ without restrictions in $I$ is when

$\frac{\mathrm{dP}}{\mathrm{dI}} = V - 2 R I = 0$ or when $I = {I}_{0} = \frac{V}{2 R}$ Putting values, for ${I}_{0} = \frac{12}{2 \times 0.5} = 12$

So ${I}_{0} \in \left[0 , 15\right]$ then it is a feasible value for $I$. This operation point represents an optimal maximum point because

$\frac{{d}^{2} P}{{\mathrm{dI}}^{2}} = - 2 R < 0$

Feb 22, 2017

$12 A$

#### Explanation:

Given: Power output of a battery  P = V I − RI^2 ......(1)
where $V$ is the emf of battery, $R$ is its internal resistance, and $I$ is the current flowing through the battery.

To find maximum power we differentiate with respect to current and set it equal to zero.
P' = d/(dI)(V I − RI^2)
V − 2RI=0#
$I = \frac{V}{2 R}$ ......(2)*
Now $P ' ' = \frac{d}{\mathrm{dI}} \left(V - 2 R I\right)$
$\implies P ' ' = - 2 R$
Since it is $- v e$ quantity, hence Power is maximum at the value of current calculated in (2)

For the given values
$I = \frac{12}{2 \times 0.5} = 12 A$

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*An important result in electrical engineering.

Power transfer between a voltage source and an external load is maximum when the internal resistance of the voltage source matches resistance of the load.