A 3.0-uF capacitor and a 4.0-uF capacitor are connected in series across a 40.0-V battery. A 10.0-uF capacitor is also connected directly across the battery terminals. Find the total charge that the battery delivers to the capacitors.
I NEED HELP WITH THIS PROBLEM. THANKS!
If I understand the circuit, the 3.0 and 4.0 uF capacitors are connected in series with a 10.0 uF capacitor connected in parallel and all of this across a 40 v cell. First determine the capacitance of the series 3.0 and 4.0.
1/C = 1/3.0 + 1/4.0 = 1.7 uF.
Then add in the parallel 10.0 to make a total of 11.7 uF.
Q = Coulombs of stored electricity = CV where C is capacitance (in F) and V is voltage. Check my thinking.
For the 10 uF capacitor,
Q' = C' V = (10*10^-6)*40 = 400 uC
For the series 4 uF and 3 uF capacitors, which both acquire the same charge Q",
10 V = Q"/4*10^-6 + Q"/3*10^-6
= Q"/1.714*10^-6
The two series capacitors act like a single 1.714 uF capacitor
Q" = 17.14 uC is the charge on both the 3 uF and 4 uF capacitors
I forgot to use 40 V for the series capacitors. Their charge is 4 times higher than what I stated. Also, the battery only has to deliver a charge of Q", not the sum of the charges on the two series capacitors, which would be 2Q"
DrBob222 already provided a correct answer by computing an effective C for all three capacitors in series-parallel
To find the total charge that the battery delivers to the capacitors, we need to understand how charges distribute in a series circuit.
In a series circuit, the same current passes through each component, but the charge on each capacitor may be different. We can calculate the total charge using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
Let's break down the problem step-by-step:
1. Capacitors in series: When capacitors are connected in series, the total capacitance (C_total) can be found using the formula: 1/C_total = 1/C1 + 1/C2 + 1/C3 + ..., where C1, C2, C3, ... are the individual capacitances. In this case, we have two capacitors (C1 = 3.0 uF and C2 = 4.0 uF), so we can calculate the total capacitance (C_total) as follows:
1/C_total = 1/3.0 uF + 1/4.0 uF
1/C_total = (4.0 + 3.0)/(3.0 x 4.0) uF
1/C_total = 7.0/(3.0 x 4.0) uF
1/C_total = 7/12 uF
C_total = 12/7 uF
C_total ≈ 1.714 uF (rounded to three decimal places)
2. Charging voltage: The voltage across the capacitors connected in series is the same as the battery voltage, which is 40.0 V in this case.
3. Calculate the total charge (Q_total): Now, we can use the formula Q = CV to find the charge on the capacitors. Since we have the total capacitance (C_total) and the voltage (V), we can plug these values into the formula:
Q_total = C_total x V
Q_total ≈ 1.714 uF x 40.0 V
Q_total ≈ 68.56 uC
Therefore, the total charge that the battery delivers to the capacitors is approximately 68.56 microCoulombs.