A capacitor with a capacitance of C = 25.7 μF is slowly charged by a constant current of I = 45.7 nA. How long does it take to charge the capacitor to a voltage of V = 29.4 V?
To calculate the time it takes to charge a capacitor, we can use the formula:
t = Q / I
Where:
t = time (in seconds)
Q = charge (in coulombs)
I = current (in amperes)
To find the charge Q, we can use the formula:
Q = C * V
Where:
Q = charge (in coulombs)
C = capacitance (in farads)
V = voltage (in volts)
Given:
C = 25.7 μF = 25.7 * 10^(-6) F
I = 45.7 nA = 45.7 * 10^(-9) A
V = 29.4 V
First, let's calculate the charge Q:
Q = C * V
Q = (25.7 * 10^(-6)) * 29.4
Q = 0.00075438 C
Now, we can calculate the time t:
t = Q / I
t = 0.00075438 / (45.7 * 10^(-9))
t = 16,499.34 seconds
Therefore, it takes approximately 16,499.34 seconds to charge the capacitor to a voltage of 29.4 V.