The resistor in an RC circuit has a resistance of 155 \Omega.
a)What capacitance must be used in this circuit if the time constant is to be 3.7 ms?
b) Using the capacitance determined in part (a), calculate the current in the circuit 7.4 ms after the switch is closed. Assume that the capacitor is uncharged initially and that the emf of the battery is 9.0 {\rm V}.
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The time constant for a resistor capacitor combination is given by
T = R x C
T = the time constant in seconds
R = the resistor value in Ohms
C = the capacitor value on Farads
so T = 3.7mS as given in the question, be careful with units 3.7 milliseconds = 3.7/1000 = 0.0037 seconds.
the resistance is 155 Ohms
so T = R x C
= 0.0037 = 155 x C
C = 0.0037/155 = 2.387 x 10 (to the power of -5) Farads.
A Farad is a very large value of capactance so most capacitors have much smaller values.
One common value of capacitance is in uF (micro Farads) i.e. 1 millionth of a Farad.
So the above value would be 23.87 uF (micro Farads).
The next bit is abit more involved, just realise 7.4mS is twice 3.7mS.
I hope this helps