time then move the Switch from \"a\" to \"b\" When the switch is in position \"b
ID: 1535593 • Letter: T
Question
time then move the Switch from "a" to "b" When the switch is in position "b", what is the time constant, r, of the circuit? Tau = R_2C tau = R_1/C tau = R_1 + R_2/R_1R_2 C tau = R_1R_2/R_1 + R_2 C tau = (R_1 + R_2)C tau = R_1 + R_2/R_1R_2 C tau = R_1C tau = R_2/C tau = (R_1 + R_2)/C tau = R_1R_2/(R_1 + R_2)C Let: V_o = 6V, C = 6 mu F, R_1 = 8 ohm, and R_2 = 18 ohm. At a time 3/2 tau after S has been switched to position "b", what is the power consumption of the circuit? Answer in units of W. Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by "leaking" across the dielectric. If a certain 5.52 mu F capacitor leaks charge such that the potential difference decreases to half its initial value in 3.8 s, what is the equivalent resistance of the dielectric? Answer in units of ohm. Your toaster oven and coffeemaker each dissipate 1200 W of power. You have a 111 V line in your kitchen. For what current must the circuit breaker be rated for you to operate both of these appliances at the same time? Answer in units of A.Explanation / Answer
given C = 5.52 micro F
let R is the resistance.
Time constant, T = R*C
let Vo is the initial potential difference across the cpacitor plates.
after t = 3.8s, V = Vo/2
we know, at time t = t,
V = Vo*e^(-t/T)
Vo/2 = Vo*e^(-t/T)
0.5 = e^(-t/T)
ln(0.5) = -t/T
T = -t/ln(0.5)
= -3.8/ln(0.5)
= 5.482
R*C = 5.482
R = 5.482/C
= 5.482/(5.52*10^-6)
= 9.93*10^5 ohms or 0.993 M ohms
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