Question #200c2

2 Answers
Feb 22, 2017

Answer:

#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#

Feb 22, 2017

Answer:

#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.