Three are used to power a portable electronic device identical batteries with E = 2.5 V each and negligible internal resistances,

connected in series. If inadvertently one of the batteries was

mounted with reverse polarity, what is the value of the supplied voltage

system?

Explanation:

Here two internal resistances are in parallel so equivalent internal resistance r

eq

=1∣∣1=

1+1

1×1

=1/2=0.5Ω

According to maximum power theorem the power will be max when internal resistance is equal to external resistance. since, r

eq

=R=0.5Ω

Maximum current through R is I

max

=

r

eq

+R

V

=

0.5+0.5

2

=2A

Maximum power is P

max

=I

max

2

R=2

2

×0.5=2W

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answer: question1

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question 3

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step-by-step explanation: