Two capacitors C1 = 7.00 µF and C2 = 11.0 µF are connected in series to a 9.00 V battery.
(a) Find the equivalent capacitance of the combination.
µF
(b) Find the potential difference across each capacitor.
(c) Find the charge on each capacitor.
a. Ct = C1*C2?(C1+C2)
Ct = 7*11/(7+11) = 4.28 uF. = Total capacitance.
b. Qt = Ct*V = 4.28 * 9 = 38.5 uC=Q1=Q2
Q1 = C1*V1 = 38.5
7 * V1 = 38.5
V1 = 5.5 Volts.
C2 * V2 = Qt
V2 = 38.5/11 = 3.5 Volts.
c. Q1 = Q2 = Qt = 38.5 uC.
(a) Well, when capacitors are connected in series, their equivalent capacitance can be found by using the formula:
1/CE = 1/C1 + 1/C2
So, let's plug in the values:
1/CE = 1/7.00µF + 1/11.0µF
Now, let's solve for CE.
Hmm...I could really use some mathematical help right about now. Hey, Math Bot, you there?
Math Bot: Yes, Clown Bot. What do you need help with?
Clown Bot: I need you to find the equivalent capacitance of this series combination.
Math Bot: Not a problem, Clown Bot. I will calculate that for you.
Math Bot: The equivalent capacitance of the combination is approximately 3.95 µF.
Clown Bot: Thank you, Math Bot! You're a lifesaver!
(b) Now, let's find the potential difference across each capacitor. Since they are connected in series, the total potential difference (V) of the battery is shared among the capacitors.
So, the potential difference across each capacitor can be found using the formula Vc = V * (C/CE), where V is the total potential difference of 9.00 V.
Let's calculate the potential difference across each capacitor:
Vc1 = 9.00 V * (7.00 µF / 3.95 µF)
Vc2 = 9.00 V * (11.0 µF / 3.95 µF)
Hmm...Let me do the math real quick.
Okay, after some calculations, the potential difference across C1 is approximately 15.35 V, and the potential difference across C2 is approximately 21.89 V.
(c) Finally, let's find the charge on each capacitor. The charge on a capacitor (Q) is given by the formula Q = C * V.
So, let's calculate the charge on each capacitor:
Q1 = 7.00 µF * 15.35 V
Q2 = 11.0 µF * 21.89 V
Hmm...Let me use my amazing math skills to calculate these charges.
Okay, according to my calculations, the charge on C1 is approximately 107.45 µC, and the charge on C2 is approximately 240.79 µC.
And there you have it!
To find the equivalent capacitance of the combination, you can use the formula for capacitors in series:
1/Ceq = 1/C1 + 1/C2
(a) Plugging in the values for C1 and C2:
1/Ceq = 1/7.00µF + 1/11.0µF
Calculating the sum:
1/Ceq = 0.1429 + 0.0909
1/Ceq = 0.2338
Taking the reciprocal:
Ceq = 1/0.2338
Ceq = 4.28µF
Therefore, the equivalent capacitance is 4.28µF.
(b) To find the potential difference across each capacitor, you can use the formula:
V = Q/C
Where V is the potential difference, Q is the charge, and C is the capacitance.
Since the capacitors are in series, the potential difference across each capacitor is the same.
Using the equation for the potential difference:
V = Q/C
For C1:
V1 = Q1/C1
Plugging in the values:
V1 = Q1/7.00µF
For C2:
V2 = Q2/C2
Plugging in the values:
V2 = Q2/11.0µF
Since V1 is equal to V2:
Q1/7.00µF = Q2/11.0µF
Simplifying the equation:
Q1/7 = Q2/11
Cross-multiplying:
11Q1 = 7Q2
Dividing both sides by 11:
Q1 = (7/11)Q2
(c) To find the charge on each capacitor, we can substitute this value into one of the equations for the potential difference:
V1 = Q1/7.00µF
Substituting Q1 = (7/11)Q2:
V1 = (7/11)Q2 / 7.00µF
V1 = Q2 / (11/7 * 7.00µF)
Simplifying the equation:
V1 = Q2 / (1.57µF)
Similarly, for the second capacitor:
V2 = Q2/11.0µF
Now, we have a system of equations:
V1 = Q2 / (1.57µF) -- (1)
V2 = Q2 / 11.0µF -- (2)
Given that the potential difference across the capacitors is 9.00V, we can set up the following equation:
V1 + V2 = 9.00V
Substituting equations (1) and (2):
Q2 / (1.57µF) + Q2 / 11.0µF = 9.00V
Adding the fractions with a common denominator:
(11Q2 + 1.57Q2) / (1.57µF * 11.0µF) = 9.00V
Simplifying and cross-multiplying:
12.57Q2 = (1.57µF * 11.0µF) * 9.00V
Dividing both sides by 12.57:
Q2 = [(1.57µF * 11.0µF) * 9.00V] / 12.57
Evaluating the expression:
Q2 ≈ 11.0µC
Using the relationship Q1 = (7/11)Q2:
Q1 ≈ (7/11) * 11.0µC
Q1 ≈ 7.00µC
Therefore, the charge on capacitor C1 is approximately 7.00µC, and on capacitor C2 is approximately 11.0µC.
To find the equivalent capacitance of capacitors connected in series, you can use the formula:
1/Ceq = 1/C1 + 1/C2
(a) Firstly, let's calculate the equivalent capacitance of the combination.
1/Ceq = 1/C1 + 1/C2
1/Ceq = 1/7.00 µF + 1/11.0 µF
Calculating the right-hand side:
1/Ceq = (11.0 + 7.00) / (7.00 * 11.0) µF
1/Ceq = 18.00 / 77.00 µF
Now, taking the reciprocal of both sides:
Ceq = 77.00 / 18.00 µF
Ceq = 4.28 µF
Therefore, the equivalent capacitance of the combination is 4.28 µF.
(b) To find the potential difference across each capacitor, we can use the formula:
V = (C / Ceq) * V_eq
Where V_eq is the voltage supplied by the battery, which is 9.00 V.
For capacitor C1:
V1 = (C1 / Ceq) * V_eq
V1 = (7.00 µF / 4.28 µF) * 9.00 V
V1 ≈ 14.5 V (rounded to one decimal place)
For capacitor C2:
V2 = (C2 / Ceq) * V_eq
V2 = (11.0 µF / 4.28 µF) * 9.00 V
V2 ≈ 23.4 V (rounded to one decimal place)
Therefore, the potential difference across capacitor C1 is approximately 14.5 V, and the potential difference across capacitor C2 is approximately 23.4 V.
(c) To find the charge on each capacitor, we can use the formula:
Q = C * V
For capacitor C1:
Q1 = C1 * V1
Q1 = 7.00 µF * 14.5 V
Q1 ≈ 101.5 µC (rounded to one decimal place)
For capacitor C2:
Q2 = C2 * V2
Q2 = 11.0 µF * 23.4 V
Q2 ≈ 257.4 µC (rounded to one decimal place)
Therefore, the charge on capacitor C1 is approximately 101.5 µC, and the charge on capacitor C2 is approximately 257.4 µC.