Question
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.
Answers
Henry
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.
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.