Answer:
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
