MasteringPhysics: Ch 24 HWa Capacitance - le Chrome https://session.masteringphy

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MasteringPhysics: Ch 24 HWa Capacitance - le Chrome https://session.masteringphysics.com/myct/.temView?assignmentProblemID=58012538&offset-prev; Reinemer202w16 Signed in as Nicolay Pekhotin HelpClose HWa Capacitanc Problem 24.14 Resources previous | 12 of 18 | next » Problem 24.14 Find the charge q1 on the capacitor C1 n the circuit shown in the figure (Figure 1), the applied potential is Vab -19.0 V.For G-2.60 F . C2 = 4.90 . and C3 = 6.65 F Submit My Answers Give Up Part B Find the charge q2 on the capacitor C2 4.9.105 Submit My Answers Give Up Incorrect, Try Again, attempts remaining Part C Find the charge q3 on the capacitor C3 Figure 1 of 1 (l Submit My Answers Give Up Part D Calculate the potential difference across the capacitor C1 10 9:08 PM 2/3/2016

Explanation / Answer

let Q1,Q2 and Q3 be the charges on capacitors C1, C2 and C3 respectively.and Ceq be the equivalent of C1 and C 2

C1 and C2 are in parallel connection

Therefore Ceq = C1 + C2 = 2.6 F + 4.9 F = 7.5 F

Ceq and C3 are in series .

Therefore C total = Ceq x C3 / ( Ceq + C3 ) = 7.5 x 6.65 / ( 7.5 + 6.65 ) = 3.524735 F

Vab = 19V

Q = V x C

Qtotal = Vab x Ctotal = 19 x 3.524735 = 66.969964 C

Since Ceq and C3 are in series, the same charge Qtotal will be on C3 and Ceq

therefore charge on C3 = Q3 = 66.969964 C = 66.969964 x 10-6 C

we know that Q = V x C

therefore, Q3 = V3 x C3

V3 = Q3 / C3 = 66.969964 x 10-6 / 6.65 x 10-6  = 10.07067 V

V3 = 10.07067 V

So the remaining voltage difference across Ceq ( or C1 or C2 , sice they are in parallel connection ) is Vab - V3

V1 = V2 = Vab - V3 = 19 - 10.07067 = 8.92933 V

V1 = 8.92933 V

Q = V x C

So, Q1 = V1 x C1 = 8.92933 x 2.6 x 10-6 = 23.216258 x 10-6 C

Therefore Q1 = 23.216258 x 10-6 C or 23.216258 C

Q2 = V2 x C2 = 8.92933 x 4.9 x 10-6 = 43.753717 x 10-6 C

Therefore Q2 = 43.753717 x 10-6 C or 43.753717 C

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