Final answer:
In a series circuit with a resistor, a capacitor takes time to charge fully. The time it takes to charge the capacitor fully is calculated using the formula t = RC, where R is the resistance and C is the capacitance. The differential equation that determines the charge as a function of time is given by dQ/dt + Q/RC = V/RC. The particular solution to the equation is Q = Vt/RC. It takes one time constant to charge the capacitor to half its capacity, which can be calculated using the formula t = RC ln(2). The current through the circuit does not change with time.
Step-by-step explanation:
(a) Time to charge the capacitor fully:
The time it takes to charge a capacitor fully when connected in series with a resistor can be found using the formula:
t = RC
where t is the time in seconds, R is the resistance in ohms, and C is the capacitance in farads.
Given that C = 15uF = 15 x 10-6 F and R = 51 ohms:
t = (15 x 10-6 F)(51 ohms) = 7.65 x 10-4 s
Therefore, it would take approximately 7.65 x 10-4 s to charge the capacitor fully.
(b) Differential equation:
The differential equation that determines the charge as a function of time is given by:
dQ/dt + Q/RC = V/RC
where dQ/dt represents the rate of change of charge with respect to time, Q is the charge on the capacitor, R is the resistance, C is the capacitance, and V is the voltage of the battery.
(c) Particular solution:
The particular solution to the differential equation can be found by assuming that the charge on the capacitor (Q) increases linearly with time. Therefore, we can write:
Q = Vt/RC
where Q is the charge on the capacitor, V is the voltage of the battery, t is the time in seconds, R is the resistance in ohms, and C is the capacitance in farads.
(d) Time to charge the capacitor to half its capacity:
A capacitor charges to half its capacity after a time interval equal to one time constant, which can be calculated using the formula:
t = RC ln(2)
Substituting the given values, we get:
t = (15 x 10-6 F)(51 ohms) ln(2) ≈ 1.56 x 10-4 s
Therefore, it would take approximately 1.56 x 10-4 s to charge the capacitor to half its total capacity.
(e) Current as a function of time:
The current through the series circuit can be calculated using Ohm's Law:
I = V/R
where I is the current, V is the voltage of the battery, and R is the resistance.
In this case, the current remains constant over time as the capacitor charges, so the current does not change with time.