Question
The label on a portable radio recommends the use of rechargeable nickel-cadmium cells (nicads), although they have a 1.25-V emf while alkaline cells have a 1.58-V emf. The radio has a $3.20\textrm{ }\Omega$ resistance. (a) Draw a circuit diagram of the radio and its batteries. Now, calculate the power delivered to the radio. (b) When using Nicad cells each having an internal resistance of $0.0400\textrm{ }\Omega$. (c) When using alkaline cells each having an internal resistance of $0.200\textrm{ }\Omega$. (d) Does this difference seem significant, considering that the radio’s effective resistance is lowered when its volume is turned up?
1. Please see the solution video
2. $0.476\textrm{ W}$
3. $0.691\textrm{ W}$
4. As the resistance decreases the power dissipated with either battery decreases as well, to the point where as the resistance approaches zero, the power dissipated in each case also approaches zero. This means the difference between the power dissipated in each case also approaches zero.

# OpenStax College Physics for AP® Courses, Chapter 21, Problem 22 (Problems & Exercises)

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