Question

A capacitor that is initially uncharged is connected in series with a resistor and a 407.1 Vemf source with negligible internal resistance. Just after the circuit is completed, the current through the resistor is 0.850 mAand the time constant for the circuit is 6.20 s. Part A: What is the resistance of the resistor?

Part B: What is the capacitance of the capacitor

Answer 1

Resistance,
`R = 478.9 k\ohm`

`C = 12.95\micro F`

Given:

V = 407.1 V

Current through the resistor, I = 0.850 mA =
`0.850* 10^(-3)`

Time constant for the circuit,
`\tau = 6.20 s`

Solution:

Initially the capacitor is uncharged, the current in the circuit remains same.

Thus

At t = 0

the current in the circuit, I =
`(V)/(R)`

Therefore,

Resistance, R =
`(V)/(I)`

R =
`(407.1)/(0.850* 10^(-3)) = 478.9 k\ohm`

Now, for calculation of Capacitance, C:

Time constant for the circuit,
`\tau = CR`

`6.20 = C* 478.9* 10^(3)`

`C = 1.295* 10^(- 5) F = 12.95\micro F`