Answer:
The current is 0.951 A
Solution:
As per the question:
Capacitance of the capacitor, C = [tex]1.70\mu F[/tex]
Resistance of the circuit, R = 14.0[tex]\Omega[/tex]
Voltage of the battery, V = 10.0 V
Now, when the charge become one-fourth of its initial value, the amount of current flowing is given by:
[tex]q' = q(1 - e^{-\frac{t}{\tau}})[/tex] (1)
where
q' = [tex]\frac{q}{4}[/tex]
[tex]\tau = RC[/tex]
[tex]q' = q(1 - e^{-\frac{t}{RC}})[/tex]
[tex]\frac{q}{4} = q(1 - e^{-\frac{t}{RC}})[/tex]
[tex]0.25 = (1 - e^{-\frac{t}{RC}})[/tex]
[tex]0.75 = e^{-\frac{t}{RC}[/tex]
[tex]-\frac{t}{RC} = ln(0.75)[/tex]
[tex]-\frac{t}{14\times 1.70\times 10^{- 6}} = - 0.2876[/tex]
[tex]t = 6.844\times 10^{- 6} s[/tex]
Now,
The current is given by:
[tex]I' = Ie^{-\frac{t}{RC}}[/tex] (2)
Also, [tex]I = \frac{V}{R} = \frac{10}{14} = 0.714 A[/tex]
Now, using the value of I = 0.714 A
[tex]I' = 0.714e^{-\frac{6.844\times 10^{- 6}}{14\times 1.70\times 10^{- 6}}} = 0.951 A[/tex]
