Consider the combination of capacitors in the figure below. (Let C1 = 27.0 mu F
ID: 1404954 • Letter: C
Question
Consider the combination of capacitors in the figure below. (Let C1 = 27.0 mu F and C2 = 3.00 mu F.) (a) Find the equivalent single capacitance of the two capacitors in series and redraw the diagram (called diagram 1) with this equivalent capacitance. (b) In diagram 1 find the equivalent capacitance of the three capacitors in parallel and redraw the diagram as a single battery and single capacitor in a loop. (c) Compute the charge on the single equivalent capacitor. (d) Returning to diagram 1, compute the charge on each individual capacitor. 4.00-muF capacitor muC 3.00-muF capacitor muC 27.0-muF and 8.00-muF equivalent capacitor muC Does the sum agree with the value found in part (c)? Yes No (e) What is the charge on the 27.0-muF capacitor and on the 8-muF capacitor? (f) Compute the voltage drop across the 27.0-muF capacitor. (g) Compute the voltage drop across the 3.00-muF capacitor.Explanation / Answer
(a)
from the circuit C1 and 8 uF are connected in series the net capacitence is
C= 27.0 * 8.0/ (27.0 + 8.0) = 6.17 uF
.
(b) Now C,C2 and 4 uF are connected in parallel
total in parallel = 4.0 + 3.00 + 6.17 = 13.17 uF
.
(C) the charge on the single equivalent capacitor is
q = CV = 13.17 *36 = 474.12uC
.
(d) the charge on the 4 uF capacitotr is
q = 4.0 *36 = 144 uC
.the charge on the 3 uF capacitotr is
q= 3.0 *36 = 108 uC
the charge on the 6.17 uF capacitotr is
q= 6.17 * 36 = 222.12uC
.
YES
.
(e)
the charge on the 27uF, 8uF capacitotr is
q= 6.17 * 36 = 222.12uC
.
(f) voltage across the 27 uF capacitor is
V = q / C = 222.12/ 27 = 8.22V
.
(g) voltage across the 3 uF capacitor is
V = q / C = 36 /3.00 = 12 V