1. A 25.0-Ohm light bulb is connected across the terminals of a 12.0-V battery h

Question

1. A 25.0-Ohm light bulb is connected across the terminals of a 12.0-V battery having 3.50 Ohm of internal resistance. a) What is current in the circuit? b) What is the rate of conversion of chemical energy to electrical energy within the battery (power output)? c) What is the rate of dissipation of critical energy in the battery due to internal resistance (power input)? d) What is the net power output of the battery? e) What is the rate of dissipation of electrical energy in the light bulb (power input)?

Explanation / Answer

a)

Here , as 25 Ohm and 3.5 Ohm are in series

Req = 25 + 3.5

Req = 28.5 Ohm

Now , using ohm's law

I = 12/28.5

I = 0.421 A

the current in the circuit is 0.421 A

b)

Pout = V*I

Pout = 12 * 0.421

Pout = 5.05 W

the power output of the battery is 5.05 W

c)

Pr = I^2*r

Pr = 0.421^2 * 3.5

Pr = 0.62 W

the power dissipated in internal resistance is 0.62 W

d)

Pnet = Pout - Pr

Pnet = 5.05 - 0.62

Pnet = 4.43 W

e)

Pin = Pnet

Pin = 4.43 W

the power dissipated in the bulb is 4.43 W

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