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.
C = q/V
so V = q/C
so
q1/C1 + q2/C2 + q3/C3 = 58
but q1=q2=q3
becaue where else are the electrons to come from or go to?
so
1/C1 + 1/C2 + 1/C3 = 58/ q
10^6 (1/3.83+1/5.89+1/12.9 = 58 /q
10^6 (.508) = 58 / q
q = (58/.508)10^-6
V = (58/.508)/3.83 = 29.8 volts
To calculate the voltage across the 3.83 μF capacitor, we need to first determine the total capacitance of the series combination of capacitors.
In a series circuit, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of individual capacitances. Mathematically, it can be expressed as:
1/C_total = 1/C1 + 1/C2 + 1/C3
Let's plug in the values:
1/C_total = 1/3.83μF + 1/5.89μF + 1/12.9μF
Calculating the individual terms:
1/C_total = 0.26075 + 0.16984 + 0.07752
Adding these values:
1/C_total ≈ 0.50811
Now, taking the reciprocal on both sides:
C_total ≈ 1/0.50811
C_total ≈ 1.9679μF
The total capacitance (C_total) is approximately 1.9679 μF.
Next, we'll use the concept of voltage division to find the voltage across the 3.83 μF capacitor.
In a series circuit, the voltage across each component is distributed in proportion to its capacitance. Mathematically, it can be expressed as:
V1/V_total = C1/C_total
Let's plug in the values:
V1/58.0V = 3.83μF / 1.9679μF
Simplifying the equation:
V1/58.0V ≈ 1.947
Now, solving for V1 (voltage across the 3.83 μF capacitor):
V1 ≈ 58.0V * 1.947
V1 ≈ 112.726V
Therefore, the voltage across the 3.83 μF capacitor is approximately 112.726V.