A 1 mF, a 2 mF, and a 3 mF capacitor are connected in series, the combination being connected across a 9 volt battery. Which capacitor has the greatest voltage?
1 mF
C = charge per unit voltage
so
V = Q/C
Like the problem below this one, since they are in series, only One charge, call it Q, comes out of the battery and lands on the top of the first capacitor. That draws a -Q from the next one which leaves +Q on the top etc.
In the end, each capacitor has charge Q
since V = Q/C
the smallest C has the biggest V
To find out which capacitor has the greatest voltage, we need to understand how capacitors behave when connected in series.
When capacitors are connected in series, the total capacitance (Ct) is given by the reciprocal of the sum of the reciprocals of the individual capacitances (C1, C2, C3):
1/Ct = 1/C1 + 1/C2 + 1/C3
Given that C1 = 1 mF, C2 = 2 mF, and C3 = 3 mF, we can substitute these values into the equation:
1/Ct = 1/1 + 1/2 + 1/3
Calculating this equation, we find:
1/Ct = 3/6 + 2/6 + 2/6
= 7/6
Taking the reciprocal of both sides, we get:
Ct = 6/7
Since the total capacitance is 6/7 mF, we can calculate the voltage across each capacitor using the formula:
V = Q/C
Where V is the voltage, Q is the charge, and C is the capacitance.
Since all three capacitors are connected in the same series circuit, the charge on each capacitor will be the same. Therefore, the voltage across each capacitor will be inversely proportional to its capacitance.
Let's calculate the voltage for each capacitor using the total charge (Q) and the reciprocal of the capacitance (1/C):
Voltage across C1 = Q / C1 = Q / (1/1) = Q / 1 = Q
Voltage across C2 = Q / C2 = Q / (1/2) = Q / (1/2) = 2Q
Voltage across C3 = Q / C3 = Q / (1/3) = Q / (1/3) = 3Q
Since the charge Q is the same for all three capacitors, the voltage is directly proportional to the capacitance. Hence, the capacitor with the greatest voltage is the one with the highest capacitance.
Therefore, the capacitor with the greatest voltage is C3, which has a capacitance of 3 mF.
To determine which capacitor has the greatest voltage in a series combination, we need to understand how capacitors behave when connected in series.
In a series circuit, the same current flows through all components. However, the voltage across each component can vary depending on its capacitance. When capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of individual capacitances (C1, C2, C3, ...).
1/C_total = 1/C1 + 1/C2 + 1/C3 + ...
In this case, we have three capacitors: C1 = 1 mF, C2 = 2 mF, and C3 = 3 mF. To find the total capacitance:
1/C_total = 1/1 + 1/2 + 1/3
1/C_total = 1/1 + 1/2 + 2/6
1/C_total = 3/6 + 2/6 + 2/6
1/C_total = 7/6
C_total = 6/7 mF
Now that we know the total capacitance, we can find the voltage across each individual capacitor using the formula:
Voltage across each capacitor = (Total Voltage) * (Individual Capacitance) / (Total Capacitance)
In this case, the total voltage (V_total) is 9 volts. Let's calculate the voltage across each capacitor:
Voltage across C1 = (9 V) * (1 mF) / (6/7 mF)
Voltage across C1 = (9 V) * (1/6/7)
Voltage across C1 = (9 V) * (7/6)
Voltage across C1 ≈ 10.5 V
Voltage across C2 = (9 V) * (2 mF) / (6/7 mF)
Voltage across C2 = (9 V) * (2/6/7)
Voltage across C2 = (9 V) * (7/6)
Voltage across C2 ≈ 10.5 V
Voltage across C3 = (9 V) * (3 mF) / (6/7 mF)
Voltage across C3 = (9 V) * (3/6/7)
Voltage across C3 = (9 V) * (7/6)
Voltage across C3 ≈ 10.5 V
From these calculations, we can see that all three capacitors have the same voltage of approximately 10.5 volts. Therefore, each capacitor in the series combination will have the same voltage across it.