Jim Guinn’s PHYS2212 Assignment #11 1. In the figure, E = 6V and R = 0.5 Ω . The rate of Joule heating in R is 8W . a) What is the current in the circuit? The Joule heating rate is P = I 2 R , thus I = (P/R) 1/2 = 4A . b) What is the potential difference across R? Similarly, P = V 2 /R or V = (PR) 1/2 = 2V . c) What is r? The voltage drop across R is 2V ; therefore, since the total voltage drop in the circuit is 6V , the voltage across r is 4V . Hence, r = V/I = 4V / 4A = 1 Ω . 2. A 0.12 μ F capacitor, in series with a resistor, is given a charge Q o . After 4s its charge is Q o /2 . What is the resistance of the resistor? It is given that Q = Q
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Voltage drop, Joule Heating, total voltage drop, Joule heating rate