Asked by Mary
The time dependence of voltage across a discharging capacitor is given by the exponential decay. The voltage on the capacitor at time t= 0 sec is 24 volt. If a 220 +/- 22uF capacitor and 1 Mega ohm (5% tolerance) resistor are used in the RC circuit, what would be the value of the time constant (T=RC) and the associated error? What would be the voltage V after t = 100 sec? Propagate the errors and find the percent error in the voltage V.
Answers
Answered by
Damon
C = q/V
q = C V
i = -dq/dt = -C dV/dt
but V = i r
so
i = -C r di/dt
i + C r di/dt = 0
V/r + C r (1/r) dV/dt) = 0
V + r C dV/dt = 0
dV/dt = -1/rC V
dV/V = -1/rC dt
ln V = - (1/rC) t
V = Vi e^-(1/RC)t
q = C V
i = -dq/dt = -C dV/dt
but V = i r
so
i = -C r di/dt
i + C r di/dt = 0
V/r + C r (1/r) dV/dt) = 0
V + r C dV/dt = 0
dV/dt = -1/rC V
dV/V = -1/rC dt
ln V = - (1/rC) t
V = Vi e^-(1/RC)t
Answered by
Damon
let T = RC
then
V = Vi e^-t/T
if RC = 10^6 * 220*10^-6 = 220 = T
Vi = 24
V = 24 e^-t/220
if t = 100
V = 24 e^-(100/220) = 24 e^-.454
= 24*.635 = 15.2
You can do the change for new C and R and get error in V
then
V = Vi e^-t/T
if RC = 10^6 * 220*10^-6 = 220 = T
Vi = 24
V = 24 e^-t/220
if t = 100
V = 24 e^-(100/220) = 24 e^-.454
= 24*.635 = 15.2
You can do the change for new C and R and get error in V
Answered by
henry2,
A. T = RC = 1*10^6 * 220*10^-6 = 220 s.
tolerance = 10 + 5 = +- 15 %.
T = 0.85*220 = 187 s, min.
T = 1.15*220 = 253 s, max.
B. t/T = 100/220 = 0.45.
V = 24/e^0.45 = 15.2 volts.
C. t/T = 100/253 = 0.395, min.
V = 24/e^0.395 = 16.2 volts, max.
t/T = 100/187 = 0.535, max.
V = 24/e^0.535 =
tolerance = 10 + 5 = +- 15 %.
T = 0.85*220 = 187 s, min.
T = 1.15*220 = 253 s, max.
B. t/T = 100/220 = 0.45.
V = 24/e^0.45 = 15.2 volts.
C. t/T = 100/253 = 0.395, min.
V = 24/e^0.395 = 16.2 volts, max.
t/T = 100/187 = 0.535, max.
V = 24/e^0.535 =
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