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dow Help 7856 Mon 7:43 PM Q E edugen willeyplus.com Wiley PLUS Chegg study order confirmation IChegg oom Mail. ThLg,7aRahoumali com FULL SCREEN PRINTER VERSION BACK NEXT Chapter 27, Problem 080 In the figure R 5.10 n, R2 10.30 n, R 15.29 n, C 5.04 HF C2 10.10 uF and the ideal battery has emf a 20.8 v. Assuming that the circuit is in the steady state, what is the total energy stored in the two capacitors? Number the tolerance is +/-1 the 3rd significant digit Click if you would like to show work for this question Open Show Mork Question Attempts o of 5 used SAvr OR umn SUBMIT ANSWER

Explanation / Answer

In steady state, both Capactiros behave like open ckt

So resistances will be connected in series

thus

RNet = R1 +R2 + R2

Rnet = 5.1 + 10.30 + 15.29

Rnet = 30.69 ohms

Inet = V/R ( ohms law)

Inet = 20.8/30.69 = 0.677 Amps

Volatge across AB = Vab = IR1

Vab = 0.677 * 5.1

Vab = 3.45 Volts

---------------------------

Voltage across ac = Vac = I R2

Vac = 0.677 * 10.30

Vac = 6.97 Volts

-----------------------------

Energy Stored across C1, U1 = 0.5 C1 Vab^2

U1 = 0.5 * 5.04 *10^-6 * 3.45^2

U1 = 30 uJ

U2 , the energy across C2 = 0.5 C2 Vac^2

U2 = 0.5 * 10.10 *10^-6 * 6.97^2

U2 = 245.33 uJ

total Enegry U = U1 + U2

Unet = 30 uJ + 245.33 uJ

Unet = 275.33 uJ

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