Question

An initially uncharged capacitor is charged by a 15.0 V battery in series with a 100kΩ resistor. If the voltage across the capacitor reaches 5.0 V in 10 seconds, what is the value of its capacitance?

Answer 1

To find the capacitance, we use the formula for voltage across a charging capacitor. Substituting given values, the capacitance is approximately 20.57 μF.

Step-by-step explanation:

To calculate the capacitance of the initially uncharged capacitor in the circuit, we need to understand the charging process of a capacitor through a resistor. When a capacitor charges through a resistor, the voltage across the capacitor as a function of time is given by V(t) = V_b(1 - e^{-t/RC}), where V_b is the battery voltage, R is the resistance, C is the capacitance, and t is the time elapsed.

From the question, we're told that V(t)=5V, V_b=15V, R=100kΩ, and t=10s. Substituting these values into the formula and solving for C, we have:
5 = 15(1 - e^{-10/(100000 ⋅ C)})
5/15 = 1 - e^{-10/(100000 ⋅ C)}
1 - 5/15 = e^{-10/(100000 ⋅ C)}
2/3 = e^{-10/(100000 ⋅ C)}
Taking the natural logarithm of both sides we get:
ln(2/3) = -10/(100000 ⋅ C)
C = -10 / ln(2/3) / 100000

C = -10 / (100000 ⋅ ln(2/3))
C ≈ 20.57 × 10^{-6} F or 20.57 μF