# A 12-Volt battery is connected in series with a 3-kΩ resister and a 5-μFcapacitor. If the capacitor is initially uncharged, determine how long ittakes before the current ﬂowing through the...

A 12-Volt battery is connected in series with a 3-kΩ resister and a 5-μF

capacitor. If the capacitor is initially uncharged, determine how long it

takes before the current ﬂowing through the resistor reduces to 73% of its initial value of 4 mA.

(a) 4.72 ms (b) 12.72 ms (c) 19.64 ms

(d) 35.83 ms (e) 41.76 ms (f) None of the above.

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### 1 Answer

A 12-volt battery is connected in series with a 3-kΩ resistor and a 5-μF

capacitor. The capacitor is initially uncharged.

When a voltage V is applied across a resistor with resistance R, the current that flows through it is I = V/R.

The initial value of current flowing through the resistor is 4 mA. As the current flows the capacitor is charged and this gradually reduces the current flowing through the circuit. When the current flowing is 73% of 4 mA, the voltage across the resistor is 0.73*12 = 8.76 V. The voltage across the plates of the capacitor is 12 - 8.76 = 3.24 V.

When a capacitor charges, the voltage after time t is given by `V(t) = V*(1 - e^(-t/(RC)))` .

Here, V(t) = 3.24, V = 12, R*C = `15*10^-3` , substituting these values in the formula allows t to be calculated.

`3.24 = 12*(1 - e^(-t/(15*10^-3)))`

=> `(1 - e^(-t/(15*10^-3))) = 3.24/12`

=> `e^(-t/(15*10^-3)) = 1 - 3.24/12`

=> `-t/(15*10^-3) = ln(8.76/12)`

=> `t = -ln(8.76/12)*15*10^-3`

=> `t ~~ 4.72` ms

The current flowing through the resistor is reduced to 73% of its initial value of 4 mA after approximately 4.72 ms

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