Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 14P
To determine

The equivalent resistance of the combination of two resistances

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If two resistors with resistances R1 and R2 are connected in parallel, then the total resistance R, measured in ohms (Ω), is given by 1/R =1/R1 + 1/R2. If R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast (in Ω/s) is R changing when R1 = 80 Ω and R2 = 100 Ω?
If the battery in the circuit below is 3.53V, resistor R1=69.44ΩΩ, and resistor R2=61.16ΩΩ, what is the magnitude of the potential drop across resistor R2?  Please give your answer in units of Volts (V).
When resistors 1 and 2 are connected in series, the equivalent resistanceis 15.4 Ω. When they are connected in parallel, the equivalent resistanceis 2.70 Ω. What are (a) the smaller resistance and (b) the largerresistance of these two resistors?

Chapter 25 Solutions

Physics for Scientists and Engineers

Ch. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - Prob. 16PCh. 25 - Prob. 17PCh. 25 - Prob. 18PCh. 25 - Prob. 19PCh. 25 - Prob. 20PCh. 25 - Prob. 21PCh. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Prob. 24PCh. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - Prob. 33PCh. 25 - Prob. 34PCh. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - Prob. 44PCh. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - Prob. 47PCh. 25 - Prob. 48PCh. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53PCh. 25 - Prob. 54PCh. 25 - Prob. 55PCh. 25 - Prob. 56PCh. 25 - Prob. 57PCh. 25 - Prob. 58PCh. 25 - Prob. 59PCh. 25 - Prob. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 62PCh. 25 - Prob. 63PCh. 25 - Prob. 64PCh. 25 - Prob. 65PCh. 25 - Prob. 66PCh. 25 - Prob. 67PCh. 25 - Prob. 68PCh. 25 - Prob. 69PCh. 25 - Prob. 70PCh. 25 - Prob. 71PCh. 25 - Prob. 72PCh. 25 - Prob. 73PCh. 25 - Prob. 74PCh. 25 - Prob. 75PCh. 25 - Prob. 76PCh. 25 - Prob. 77PCh. 25 - Prob. 78PCh. 25 - Prob. 79PCh. 25 - Prob. 80PCh. 25 - Prob. 81PCh. 25 - Prob. 82PCh. 25 - Prob. 83PCh. 25 - Prob. 84PCh. 25 - Prob. 85PCh. 25 - Prob. 86PCh. 25 - Prob. 87PCh. 25 - Prob. 88PCh. 25 - Prob. 89PCh. 25 - Prob. 90PCh. 25 - Prob. 91PCh. 25 - Prob. 92PCh. 25 - Prob. 93PCh. 25 - Prob. 94PCh. 25 - Prob. 95PCh. 25 - Prob. 96PCh. 25 - Prob. 97PCh. 25 - Prob. 98PCh. 25 - Prob. 99PCh. 25 - Prob. 100PCh. 25 - Prob. 101PCh. 25 - Prob. 102PCh. 25 - Prob. 103PCh. 25 - Prob. 104PCh. 25 - Prob. 105PCh. 25 - Prob. 106PCh. 25 - Prob. 107PCh. 25 - Prob. 108PCh. 25 - Prob. 109PCh. 25 - Prob. 110PCh. 25 - Prob. 111PCh. 25 - Prob. 112PCh. 25 - Prob. 113PCh. 25 - Prob. 114PCh. 25 - Prob. 115PCh. 25 - Prob. 116PCh. 25 - Prob. 117PCh. 25 - Prob. 118PCh. 25 - Prob. 119PCh. 25 - Prob. 120P
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Ohm's law Explained; Author: ALL ABOUT ELECTRONICS;https://www.youtube.com/watch?v=PV8CMZZKrB4;License: Standard YouTube License, CC-BY