A20.0 μf capacitor initially charged to 30.0 μc is discharged through a 1.20 kω resistor. how long does it take to reduce the capacitor's charge to 15.0 μc

16.6 ms or 0.0166 s

Explanation:

If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken  and τ the time constant,

[tex]Q = Q'( 1- e^{-t\div \tau })

τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s

15 = 30 ( 1- e^{-t\div \ 0.024 })

( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30

⇒ - e^{-t\div \ 0.024 }) = 0.5 -1  

⇒  e^{-t\div \ 0.024 }) = 0.5

Taking logarithm to the base e on both sides of this equation,

⇒ t = 0.0166 seconds = 16.6 milli seconds