Consider the combination of capacitors in the figure below where C1=7.5nF, C2=18
ID: 2012582 • Letter: C
Question
Consider the combination of capacitors in the figure below where C1=7.5nF, C2=18.0nF, C3=30.0nF, C4=10.0nF, and C5=6.5nF.
a.) Find the equivalent single capacitance of the capacitors in series and redraw the diagram. Label it as figure 1.
b.) In figure 1, find the equivalent capacitance of the capacitors and redraw the diagram as a single battery and single capacitor in a loop. Label it as figure 2.
c.) Compute the charge on the single equivalent capacitor.
d.) Compute the charge on each individual capacitor in figure 1.
e.) Compute the charge on C2, C3, and C4.
Explanation / Answer
a) equivalent capacitance of capacitors in series depends on the equation Ceq = 1/(1/C2 + 1/C3 + 1/C4), so the series equivalent capacitance C234 = 1/(1/18 + 1/30 + 1/10) 5.3nF.
b) now you have three capacitors in parallel, so to find the equivalent capacitance just add them together: 7.5nF + 5.3nF + 6.5nF = 19.3nF.
c) charge Q = CV = 19.3nF * 25V = 482.5nC
d) figure 1 contains three caps in parallel (C1, C234, and C3). Caps in parallel share the same voltage, in this case 25V. so Q1 = 7.5nF * 25V = 187.5nC. Q234 = 5.3nF * 25V = 132.5nC. Q5 = 6.5nF * 25V = 162.5nC.
e) caps in series share the same charge, which you already figured out in part d was 132.5nC. So Q2 = Q3 = Q4 = 132.5nC.
f) You alread figured out the voltage drop over C1 and C5 (being parallel with 25V, they are both 25V.) Now to figure out the voltage drop across C2, C3, and C4: you know the capacitance and the charge, so to find the voltage, V = Q/C. V2 = 132.5nC/18nF = 7.36V. V3 = 132.5nC/30nF = 4.42V. V4 = 132.5nC/10nF = 13.25V.