A power supply has an open-circuit voltage of 37.0 V and an internal resistance

Question

A power supply has an open-circuit voltage of 37.0 V and an internal resistance of 2.00 . It is used to charge two storage batteries connected in series, each having an emf of 5.50 V and internal resistance of 0.300 . The charging current is to be 3.50 A.

(a) What additional resistance should be added in series?

(b) At what rate does the internal energy increase in the supply?

W

(c) At what rate does the internal energy increase in the batteries?

W

(d) At what rate does the internal energy increase in the added series resistance?

W

(e) At what rate does the chemical energy increase in the batteries?

W

Explanation / Answer

V1=37 V

r1=2 ohms

emf v2=5.5 v

r2=0.3 ohms

I=3.5 A

a)

use,

V1-2*V2=I*(R+r1+2*r2)

37-2*5.5=3.5*(R+2+2*0.3)

===> resistance, R=4.83 ohms

b)

power at power supply,

P=I^2*r1

P=3.5^2*2

P=24.5 W

c)

power at batteries,

P=I^2*2*r2

=3.5^2*(2*0.3)

=7.35 W

d)

power at resistor,

P=I^2*R

=3.5^2*4.83

=59.2 W

e)

rate of chemical energy increase,

P=2*V2*I

=2*5.5*3.5

=38.5 W

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