Question

What resistance is needed in this rc circuit if the flash is to charge to 90% of its full charge in 22 s ?

Answer 1

Missing part in the text of the problem:
"a flash unit for a camera has a capacitance of1200μF."

Solution:
In a RC circuit, the charge of the capacitor at time t follows the relationship:

`Q(t) = Q_0 (1-e^{- (t)/(\tau) })`
where
`Q_0` is the full charge, and
`\tau = RC` is the time constant of the circuit.

We can isolate
`\tau` from the previous equation:

`(Q(t))/(Q_0) = 1-e^{ (t)/(\tau) }`

`(t)/(\tau) = -ln(1- (Q(t))/(Q_0))`

`\tau = - (t)/(ln(1- (Q(t))/(Q_0) ))`

We can now using the data of the problem. We know that after a time t=22.0s, the capacitor is at 90% of tis charge, therefore
`(Q(t))/(Q_0) = 0.9`. So we find

`\tau = - (22)/(ln(1-0.9))=9.55 s`

And from this value we can find the value of the resistance R, since we know that
`\tau = RC`. Given
`C=1200 \mu F = 1200 \cdot 10^(-6) F`, we have

`R= (\tau)/(C)= (9.55s)/(1200 \cdot 10^(-6)F)=7958 \Omega = 7.96 k \Omega`