Solution Found!
A 25.0- ? bulb is connected across the terminals of a
Chapter 25, Problem 49E(choose chapter or problem)
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?
Questions & Answers
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)