A 25.0- ? bulb is connected across the terminals of a

Chapter 25, Problem 49E

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QUESTION:

Problem 49E

A 25.0- Ω bulb is connected across the terminals of a 12.0-V battery having 3.50 Ω of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

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QUESTION:

Problem 49E

A 25.0- Ω bulb is connected across the terminals of a 12.0-V battery having 3.50 Ω of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

ANSWER:

Solution 49E

Step 1:

In this problem, we have to calculate the fraction of power dissipated due to internal resistance and which is not available to the bulb.

 The power of a circuit can be defined as, P = I2 R

Then, the power lost due to internal resistance r, Pin = I2 r

The total resistance of the circuit, R’ = r + R

Then, the total power, Ptot = I2 R’ = I2 (r+R)

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