- O X B MasteringPhysics: HW #6 - Google Chrome A Secure https://session.masteri
ID: 1583841 • Letter: #
Question
- O X B MasteringPhysics: HW #6 - Google Chrome A Secure https://session.masteringphysics.com/myct/itemView?assignmentProblemlD=90215194 ( HW #6 Problem 23.55 () - Part A What is the potential difference between the terminals of the battery? Express your answer using two significant figures. Constants A real battery is not just an emf. We can model areal 1.5 Vnrm VW battery as a 1.5 { m V) emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1). A typical battery has 1.0 km Omega) internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 {vm VA), the value of the emf. Suppose the terminals of this battery are connected to a 2.4 kvm ( Omega) resistor. vm) = 1.06 {m V) Previous Answers Correct Part B Figure What fraction of the battery's power is dissipated by the internal resistance? Express your answer using two significant figures. Ya AEp 6 + O DI ? Vargefrac{Delta PXP3W = 3U (1961) 1.02 AliwTE Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining 1.5 V Provide Feedback Next > O Type here to search o e e 9 E 11:01 AM a & # D dx 3n/2018 1Explanation / Answer
a) net resistance = 1 + 2.4 = 3.4 ohms
current V / R = 1.5 / 3.4 = 0.44 A
potential difference = V - I R = 1.5 - (1 * 0.44)
potential difference = 1.06 V
b) power = 1.5 * 0.44 = 0.66 W
power dissipated [internal resistance] = I2 R = 0.442 * 1 = 0.1936 W
Fraction of batery's power dissipated through internal resistance
= 0.1936 / 0.66
= 0.293 = 29.3 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.