You have a circuit with a 4 uF capacitor connected in series to 2 capacitors connected in parallel (2 uF and 1.5 uF). The battery is 12 V.
The first question asked what the circuit's equivalence capacitance was. I calculated it to be 1.9 uF, which is correct.
The next question asked how much charge flows through the battery as the capacitors are being charged.
I went q=cdeltaV=1.9 x 12= 23uC
which is incorrect. What is the correct answer and how do you get to it?
Well, the equivalent capacitance is found, so C=q/V or q= CV. You did it correctly, however, I would have carried it to more places, giving 22.4microC
I see nothing wrong with your approach.
To correctly determine the charge that flows through the battery as the capacitors are being charged, we need to consider the principles of capacitance and how they apply in this circuit.
In a series circuit, the equivalent capacitance (Ce) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
1 / Ce = 1 / C1 + 1 / C2 + 1 / C3
Given that C1 = 4 uF, C2 = 2 uF, and C3 = 1.5 uF, we can substitute these values into the equation:
1 / Ce = 1 / 4 + 1 / 2 + 1 / 1.5
Now, let's simplify the equation:
1 / Ce = 0.25 + 0.5 + 0.67
1 / Ce = 1.42
To find Ce, we take the reciprocal of both sides:
Ce = 1 / 1.42
Ce ≈ 0.7042 uF
So, the correct equivalent capacitance of the circuit is approximately 0.7042 uF.
Now, to determine the charge that flows through the battery as the capacitors are being charged, we use the formula:
Q = C * ΔV
Where Q is the charge, C is the capacitance, and ΔV is the change in voltage across the capacitor.
Since the battery voltage is 12 V, and the equivalent capacitance is 0.7042 uF, we can substitute these values into the formula:
Q = 0.7042 μF * 12 V
Q ≈ 8.45 μC
Therefore, the correct answer is approximately 8.45 μC, which is the amount of charge that flows through the battery as the capacitors are being charged.