Question

Question #200c2

Answer 1

`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

`(dP)/(dI) = V-2R I=0` or when `I = I_0 = V/(2R)` Putting values, for `I_0 = 12/(2 xx 0.5)=12`

So `I_0 in [0, 15]` then it is a feasible value for `I`. This operation point represents an optimal maximum point because

`(d^2P)/(dI^2) = -2R < 0`

Answer 2

`12A`

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=V/(2R)` ......(2)*
Now `P''=d/(dI)(V-2RI)`
`=>P''=-2R`
Since it is `-ve` quantity, hence Power is maximum at the value of current calculated in (2)

For the given values
`I=12/(2xx0.5)=12A`

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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.