Exercise 26.42 20 of 25> Constants Part A You connect a battery, resistor, and c
ID: 1585312 • Letter: E
Question
Exercise 26.42 20 of 25> Constants Part A You connect a battery, resistor, and capacitor as in (Figure 1), where R= 17.0 and C= 9.00 × 10 6 F The switch S is closed at t = 0, when the current in the circuit has magnitude 3.00 A, the charge on the capacitor is 40.0 × 10 6 C. What is the emf of the battery? Express your answer with the appropriate units. Value Units Submit Part B At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? Express your answer with the appropriate units. t-Value Units Submit Part C When the current has magnitude 3.00 A, at what rate is energy being stored in the capacitor? Express your answer with the appropriate units. Palue Units Submit Figure Part D Switch When the current has magnitude 3.00 A, at what rate is energy being supplied by the battery? Express your answer with the appropriate units. open Value Units i=0 q= Submit Provide Feedback Next >Explanation / Answer
Given,
R = 17 Ohm ; C = 9 x 10^-6 F ; I = 3 A ; Q = 40 x 10^-6 C
A)using Kirchoff's loop rule
E - R I - Q/C = 0
E = R I + Q/C
E = 17 x 3 + 40 x 10^-6/9 x 10^-6 = 55.44 V
Hence, E = 55.44 V
B)We know that
Q = C Vb (1 - e^-t/RC)
40 x 10^-6 = 9 x 10^-6 x 55.44 (1 - e^-t/RC)
0.08 = (1 - e^-t/RC)
e^-t/RC = 1 - 0.08 = 0.92
-t/RC = ln0.92 = -0.0833
t = RC x 0.0833 = 17 x 9 x 10^-6 x 0.0833 = 12.7 us
Hence, t = 12.7 x 10^-6 s = 12.7 us
C)P = q I/C
P = 40 x 10^-6 x 3/(9 x 10^-6) = 13.33 W
Hence, P = 13.33 W
D)P = i^2 R
P = 3^2 x 17 = 153 W
P' = 153 + 13.33
Hence, P = 166.33
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