webassign.net 9.8uf And 21.2uf Capacitor Are C U2- Quiz Chap 24 2. 0.5/2 points

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webassign.net 9.8uf And 21.2uf Capacitor Are C U2- Quiz Chap 24 2. 0.5/2 points | Previous Answers Tipler6 24.P.034. For the circuit shown in the figure below, (V 11.0 V) 030 pF 0pF 25 p (a) Find the equivalent capacitance between the terminals 0.24 (b) Find the charge stored on the positively charged plate of each capacitor 0.3?F capacitor 2.64 1.00 uf capacitor 0.250 uF capacitorHC (c) Find the voltage across each capacitor 0.3 ?F capacitor 1.00 ?F capacitor 0.250 F capacitor (d) Find the total stored energy eBook Submit Answer Save Progress Practice Another Version

Explanation / Answer

Remember:

For parallel combination

Ceq = C1 + C2 + C3 +...............

for series combination

1/Ceq = 1/C1 + 1/C2 + 1/C3 + ............

for 2 capacitors in series it will be

Ceq = C1*C2/(C1+C2)

Using this Information:

C2 and C3 are in parallel, So

C23 = C2 + C3 = 1 + 0.25 = 1.25 uF

C1 and C23 are in series, So

Ceq = C1*C23/(C1 + C23) = 0.3*1.25/(0.3 + 1.25) = 0.24 uF

Now

Qeq = Ceq*V = 0.24*11 = 2.64 uC

Now remember in capacitors parallel combination voltage distribution in each part will be same and in series combination charge distribution in each capacitor will be same.

Charge in C1 = Charge in C23 = 2.64 uC

Q1 (0.30 uF) = 2.64 uC

V1 = Q1/C1 = 2.64/0.3 = 8.8 V

V23 = V - V1 = 11 - 8.8 = 2.2 V

Since C2 and C3 are in parallel, So

V2 = V3 = V23

V2 (1.0 uF) = 2.2 V

V3 (0.25 uF) = 2.2 V

Q2 = C2*V2 = 1*2.2 = 2.2 uC

Q3 = C3*V3 = 0.25*2.2 = 0.55 uC

Total energy stored = 0.5*C*V^2 = 0.5*0.24*10^-6*11^2 = 0.145 uJ

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