Three capacitors (3.83, 5.89, and 12.9 μF) are connected in series across a 58.0 V battery. Calculate the voltage across the 3.83 μF capacitor.
E = 58.0 Volts
1/Ct = 1/C1 + 1/C2 + 1/C3
1/Ct = 1/3.83 + 1/5.89 + 1/12.9 = 0.5084
Ct = 1.97 uF.
Qt = Q1 = Q2 = Q3.
Qt = Ct*E = 1.97 * 58 = 114.1 uC
Q1 = C1*V1 = 114.1 uC
V1 = 114.1/3.83 = 29.8 Volts.
To find the voltage across the 3.83 μF capacitor in this series circuit, we first need to calculate the equivalent capacitance.
The equivalent capacitance (C_eq) of capacitors in series can be calculated using the formula:
1/C_eq = 1/C1 + 1/C2 + 1/C3 + ...
In this case, we have three capacitors in series:
1/C_eq = 1/3.83 μF + 1/5.89 μF + 1/12.9 μF
To calculate the equivalent capacitance, we add the reciprocals of the individual capacitances:
1/C_eq = (1/3.83 μF) + (1/5.89 μF) + (1/12.9 μF)
Next, we can find the reciprocal of the sum:
1/C_eq = 0.2608 + 0.1699 + 0.0775 = 0.5082
Finally, we take the reciprocal of both sides to get the equivalent capacitance:
C_eq = 1 / (0.5082) = 1.967 μF (approximately)
Now that we have the equivalent capacitance, we can use it to calculate the voltage across the 3.83 μF capacitor, which is the same as the voltage across the equivalent capacitor.
The voltage across a capacitor (V) in a circuit can be calculated using the formula:
V = Q / C
Where Q is the charge on the capacitor and C is the capacitance.
Since the equivalent capacitance is the same as the capacitance of the 3.83 μF capacitor, the voltage across the 3.83 μF capacitor is the same as the voltage across the equivalent capacitance.
Using the given information, we can calculate the voltage across the equivalent capacitance:
V = Q / C_eq
Since the capacitors are connected in series, they share the same charge (Q). Therefore, the voltage across the equivalent capacitance is the same as the voltage across each individual capacitor.
V = 58.0 V
Hence, the voltage across the 3.83 μF capacitor (and the equivalent capacitance) is 58.0 V.