Question:

An electric circuit consists of a 10 Ω resistor, a 5 μF capacitor, and a 12 V battery connected in series. The switch is closed at time t=0. Calculate the time it takes for the charge on the capacitor to reach 90% of its maximum value.

Solution:

When the switch is closed, the capacitor starts to charge up. We can use the equation for the charging of a capacitor in an RC circuit to determine the time it takes for the charge on the capacitor to reach a certain percentage of its maximum value.

The equation for the charge on the capacitor as a function of time is given by:

Q(t) = Q_max * (1 - e^(-t / RC))

Where: Q(t) = Charge on the capacitor at time t Q_max = Maximum charge on the capacitor (when fully charged) t = Time R = Resistance in the circuit (10 Ω) C = Capacitance of the capacitor (5 μF)

We are given that the charge on the capacitor reaches 90% of its maximum value. Therefore, we can write:

0.9 * Q_max = Q_max * (1 - e^(-t / RC))

Simplifying the equation, we have:

0.9 = 1 - e^(-t / RC)

Rearranging the equation, we get:

e^(-t / RC) = 0.1

Taking the natural logarithm (ln) of both sides, we have:

-t / RC = ln(0.1)

Solving for t, we get:

t = -RC * ln(0.1)

Now, let's substitute the given values into the equation to calculate the time it takes for the charge on the capacitor to reach 90% of its maximum value:

R = 10 Ω
C = 5 μF = 5 * 10^(-6) F

t = -(10 Ω) * (5 * 10^(-6) F) * ln(0.1)

t ≈ 35.39 ms (rounded to two decimal places)

Therefore, it takes approximately 35.39 ms for the charge on the capacitor to reach 90% of its maximum value.