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

Answer:

Resistance, $R = 478.9 k\ohm$

$C = 12.95\micro F$

Given:

V = 407.1 V

Current through the resistor, I = 0.850 mA = $0.850\times 10^{-3}$

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 = $\frac{V}{R}$

Therefore,

Resistance, R = $\frac{V}{I}$

R = $\frac{407.1}{0.850\times 10^{-3}} = 478.9 k\ohm$

Now, for calculation of Capacitance, C:

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

$6.20 = C\times 478.9\times 10^{3}$

$C = 1.295\times 10^{- 5} F = 12.95\micro F$