Question

The time, in seconds, for the current through a 870-ohm resistor in an RC circuit, with an initially uncharged 4.24-mF capacitor, to fall to -37% of its maximum value is Add your answer Question 3 7.14 Points The time for the current in the resistor of an RL circuit, in units of ms, to reach -63% of its final value with a 705 resistor and a 84.7 mH inductor is Add your answer

Answer 1

Question 2

R = 870 Ω, C = 0.00424 F

Charge at a given time t is q = 0.37 Q

Time constant of RC circuit, T = RC = 870 x 0.000424 = 0.368 s

Instantaneous charge, q = Q { 1 – e^-t/T}

0.37 Q = Q { 1 – e^-t/T}

{ 1 – e^-t/T} = 0.37

e^-t/T = 0.63 ; e^t/T = 1.587 : t/T = ln1.587 = 0.462: t = 0.462 x T = 0.462 x 0.368 = 0.17 s

time taken, t = 0.17 s

question 3

R = 705 Ω; L = 0.0847 H

Time constant of RL circuit, T = L/R = 0.0847/705 = 0.12 x 10^-3 = 0.12 ms

In a RL circuit, the time taken for the current to become 63 % of maximum current is called time constant of the circuit T.

Time for the current to rise to 63 % of maximum current = 0.12 ms

Question 4

Resistor—current and voltage are in phase

Inductor—voltage leads current by π/2 radians

Capacitor—current leads voltage by π/2 radians