Problem 8: I have the correct answer and I\'m pretty sure I know what formula he

Question

Problem 8: I have the correct answer and I'm pretty sure I know what formula he used to solve the problem. I just dont quite understand how he got the answer. If someone could explain this to me that would be great!

Find the total capacitance for the combination of capacitors shown below

20.23 ppo(1+AT) Power g142 18.3] Coulomb's Law: F,-k 18.Eqa small "test charge" 18.13 Ek [ 19.1] v , q a small "test charge" [134] 1 Volt = 1 Joule / 1 Coulomb 20.26] P- PE gV 20.28] P-IV [20.29] P- 20.30 P-IR [20.34] Energy and Power: E-Pt Alternating Current: 20.38 V(t)-Vsin(2ft) [20.39] 1(t)=I,sin(2z/t) [ 19.13] 1 eV= 1.602176565 × 10 19.25) uniform E-field only [19.36] E_AV Capacitors [19.50] C-2 19.51] 1 Farad-1 Coulomb 1 Volt Parallel-Plate Capacitor: (20.39b R-Ya V [20.40a] p(t)=1(t)V(t)=lov,sin2Zrft) [20.40] P2,V [1953] C=ET Parallel-Plate Capacitor with Dielectric: [19.57] C=KEgd Capacitors in Series: [19.63] --+-+ [20.42] V.ma 120.461 P1 12 [20.29] Pang 230P-1 R Resistors in Series: 21.51 RRR2+Rt... Resistors in Parallel: Capacitors in Parallel: 19.69] C C+C+C... Energy Stored in Capacitor: [19.74] Et ,QV [21.44] V=ESMy-IR,ernal 121.45 IR 19.75E-19 EaMy 2 C [19.771 E-c Electric Current: [20.1] 1=AQ [20.2] 1 Ampere = 1 Coulomb / 1 second [20.8] 1=nqAv, [20.14] Ohm's Law: I= Kirchoffs Junction Rule 21.K1 I,+11,+..-0 (out from junction) Kirchoff's Loop Rule [21.K21 V+VV,0 (around loop) Galvanometer as Voltmeter At 21.681 R-R RR is large R 1 Volt [20.15] 10hm1Nz [20.171 Ohm's Law: V IR [20.18] R=PL Galvanometer as Ammeter: 21.70aIRi is small Voltage across Charging Capacitor in RC Circuit: [21.77] V(r)=11-e"cmc [21.78] =RC 1 Ampere

Explanation / Answer

if the circuit has a series capacitors then

1/Cs = 1/C1 +1/C2 + ........

and

if that capacitors are connected in parallel

Cp = C1 + C2 + ............

in the given result also used same formula and technique

that is the following way

in the given circuit

the 300 nF and 100 nF are in parallel

that is why 300 nF + 100 nF

then resultant is in series with 100 nF and 100 nF which are series

that is why ( 100nF -1 + 100nF-1 )-1

so totally

C = 300 nF + 100 nF + ( 100nF -1 + 100nF-1 )-1

C = 450 nF

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