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The hot resistance of a flashlight bulb is 2.30 ? , and it
Chapter 21, Problem 21(choose chapter or problem)
The hot resistance of a flashlight bulb is \(2.30 \Omega\) , and it is run by a \(1.58-V\) alkaline cell having a \(0.100-\Omega\) internal resistance.
(a) What current flows?
(b) Calculate the power supplied to the bulb using \(I^{2} R_{\text {bulb }}\).
(c) Is this power the same as calculated using \(\frac{V^{2}}{R_{\text {bulb }}}\)?
Equation Transcription:
Text Transcription:
2.30 Omega
1.58-V
0.100-Omega
I^2R_bulb
V^2 / R_bulb
Questions & Answers
QUESTION:
The hot resistance of a flashlight bulb is \(2.30 \Omega\) , and it is run by a \(1.58-V\) alkaline cell having a \(0.100-\Omega\) internal resistance.
(a) What current flows?
(b) Calculate the power supplied to the bulb using \(I^{2} R_{\text {bulb }}\).
(c) Is this power the same as calculated using \(\frac{V^{2}}{R_{\text {bulb }}}\)?
Equation Transcription:
Text Transcription:
2.30 Omega
1.58-V
0.100-Omega
I^2R_bulb
V^2 / R_bulb
ANSWER: