Question Details My Notes Ask Your Consider the combination of capacitors in the

Question

Question Details My Notes Ask Your Consider the combination of capacitors in the figure below (C1 = 27.0 F, C2 2.50 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 F capacitor 2.50 F capacitor 27.0-uF and 8.00-HF equivalent capacitor HC Does the sum agree with the value found in part (c)? HC HC (e) what is the charge on the 27.0 F capacitor and on the 8-uF capacitor? HC (f) Compute the voltage drop across the 27.0 F capacitor. (g) Compute the voltage drop across the 2.50 F capacitor. Need Help?

Explanation / Answer

let,

C1=27 uF, C2=2.5 uF

C3=4uF, C4=8 uF and V=36v

a)

C14=C1*C4/(C1+C4)

=27*8/(27+8)

=6.17 uF

b)

C_equ=C1234=C14+C2+C3

=6.17+2.5+4

=12.67 uF

c)

q_equi=C_equi*V

=12.67*10^-6*36

=456.12*10^-6 C

d)

here,

q3=C3*V

q3=4*10^-6*36

q3=144 uC

====> charge on C3=4uF is q3=144 uC

q2=C2*V

q2=2.5*10^-6*36

q2=90 uC

===> charge on C2=2.5 uF is q2=90 uC

q14=C14*V

q14=6.17*10^-6*36

q14=222.12 uC

===> charge on C14=2.5 uF is q14=90 uC

here,

q2+q3+q14=(90+144+222.12)*10^-6

=456.12 uC

==> q_equi=q3+q4+q14

e)

here,

q1=q4=q14=222.12uC

charge on C1=27 uF and C4=8uF is q14=222.12uF

f)

voltage drop across C1=27uF is

V1=q1/C1=222.12*10^-6/(27*10^-6)

V1=8.23 v

g)

voltage drop across C2=27uF is

V2=q2/C2=90*10^-6/(2.5*10^-6)

V2=36 v

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