RC charging circuit

A 5 F \text{F} capacitor is connected in series to a 4 V \text{V} battery, a 3 Ω \Omega resistor, and a switch. What is the charge on the capacitor, in C \text{C} , when the switch has been closed for 15 ln 10 s 15 \ln10 \text{ s} ?


The answer is 18.

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1 solution

July Thomas
Apr 19, 2016

Relevant wiki: RC Circuits (Direct Current)

The equation for the charge on the capacitor in and RC circuit is

q = q m a x ( 1 e t τ ) . q = q_{max}(1-e^{\frac{-t}{\tau}}).

Given t = 15 ln 10 s t = 15 \ln10 \text{ s} , calculate

q m a x = ε C = ( 4 V ) ( 5 F ) = 20 C q_{max} = \varepsilon C = (4 \text{V})(5 \text{F}) = 20 \text{C}

and

τ = R C = ( 3 Ω ) ( 5 F ) = 15 s . \tau = RC = (3 \Omega)(5 \text{F}) = 15 \text{s}.

Hence,

q = q m a x ( 1 e t τ ) = 20 ( 1 e 15 ln 10 15 ) = 20 ( 1 e ln 10 ) = 20 ( 1 0.1 ) = 18 C . q = q_{max}(1-e^{\frac{-t}{\tau}}) = 20(1-e^\frac{-15 \ln10}{15}) = 20(1-e^{-\ln10}) = 20(1-0.1) = 18 \text{C}.

Good solution.

Eud briel - 6 months, 3 weeks ago

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