Suppose you have the following circuit. The switch has been open for a long time, and is closed at time t = 0.Assume V, = 5V,R2 = 10kN, L2 = 2mH,C2 = 0.1µFThe second order differential equation for the voltage across the capacitor isdV.(t)/dt + a,dV.(t)/dt + a,Ve = KR2L2V2C2What is the coefficient K of the 2nd order differential equation for this circuit? Do not use scientific notation. In the circuit below, the switch is closed for a long time beforetime t = 0. V, = 24V, R1 = 300N, C1 = 104F and C, = 204F.C1R1t = 0-V.+VsC2What is the voltage V. across capacitor C, when the switch is closed? Give your answer in V, without entering theunits.

For first circuit we will apply KVL in loop to find diffrential equation relating capacitor voltage…

Answered: Suppose you have the following circuit.…

Delmar's Standard Textbook Of Electricity

7th Edition

ISBN:9781337900348

Author:Stephen L. Herman

Publisher:Stephen L. Herman

Chapter19: Capacitors

Section: Chapter Questions

Problem 2PA: You are an electrician working in an industrial plant. You discover that the problem with a certain...

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You are an electrician working in an industrial plant. You discover that the problem with a certain machine is a defective capacitor. The capacitor is connected to a 240-volt AC circuit. The information on the capacitor reveals that it has a capacitance value of 10 mF and a voltage rating of 240 VAC. The only 10-mF AC capacitor in the storeroom is marked with a voltage rating of 350 WVDC. Can this capacitor be used to replace the defective capacitor? Explain your answer.
For R1=9000, R2=2000, R3=6000, R4=5000, C1=0.006, C2=D0.002 & 11=0.007 A in the shown circuit, obtain the energy stored in each capacitor under dc conditions. C2 R4 ŽR3 ER2 ŽRI C1= Energy stored in C1= Energy stored in C2=
A capacitor of capacity C₁ is charged upto V volt and then connected in parallel to an uncharged capacitor of capacity C₂. The final potential difference across each will be (a) (c) C₂V G₁ + C₂ (1+2) GV G + C₂ (4) (1-²) v (b)
/Q6 The potential at point P due to Q1 and * :Q2 is = 1x10-12C 0.5 m 0.5 m Q2 = 1x10-12C %3D 50 cm P. 0.01797 V O -0.01271 V O ov O 0.01271 V C 10
The circuit in Figure has been connected for a long time. If the battery is disconnected, how does it take the capacitor to discharge to 1/20 of its initial voltage? Take R₁ = 2.00 Q, R₂=4.00 Q, R3-7.00 Q, R4 = 2.00 Q and C= 5.00 uF and ε = 14.0 V. (Your result must be in us. Include 2 digit after the decimal point and maximum of 2% of error is accepted in your answer.) E R₂ R₂ BA
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For R1=5000 £2, R2=8000, R3-9000, R4-8000 , C1=0.008 F, C2=0.002 F & I1=0.002 A in the shown circuit, obtain the energy stored in each capacitor under dc conditions. C2 11(4 {R3 www R4 C1 Energy stored in C1 (in Joule) = Oa. 0.033471074380165 Ob. 0.1004132231405 Oc. 0.066942148760331 Od. 0.13388429752066 Energy stored in C2 (in Joule)= Oa. 0.017137190082645 Ob. 0.085685950413223 Oc. 0.042842975206612 Od. 0.064264462809917 R2 R1
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In the figure ɛ = 10.0 V, R1 = 4.00 N, and R2 = 1.00 N. The inductor is ideal. If the switch is closed for a long time, what is the current through the inductor. Give your answer in A. Şekildeki devrede ɛ = 10.0 V, R1 = 4.00 N , ve R2 = 1.00 N olarak verilmişlerdir. Solenoidin iç direnci yoktur. Anahtar kapatıldıktan çok uzun süre sonra solenoidden geçen akım A cinsinden nedir. R1 R2 L E Yanıtınız
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