#9 please The maximum allowable potential difference across a 0.500 H inductor i

Question

#9 please

The maximum allowable potential difference across a 0.500 H inductor is 400. V. You need to raise the current through the inductor from 1.5 A to 3.3 A. What is the minimum time you should allow for the changing the current? Electric potentials are plotted on a 1.0 m times 1.0 m grid. b. Draw the 50 V, 75 V, and 100 V equipotential surfaces. c. Determine the electric field strength at the points A, B, C, and D. d. If a proton is release at point B, which way will it accelerate? (Draw arrow on diagram.) d. What is the work done by the electric field to move a 1.0-C charge from point B to point D? If the potential difference V = 100 V is applied across the capacitor arrangement with capacitances C_1 = 10.0 mu F, C_2 = 5.00 mu F, and C_3 = 15.0 mu F, what are the a) charge Q_3 b) potential difference V_3 c) stored energy U_3?

Explanation / Answer

C1 and C2 are in parallel so net capacitance wil be C1+C2=15 micro farad=Cp

now Cp and C3 are in series so1/ Cnet = 1/Cp+1/15= 2/15

Cnet=7.5micro farad

Total charge in The series combination of C3 And Cp wil be Q=Cnet ×V=7.5 ×10-6×100=7.5× 10-4C

CHARGE IN THE SERIES WILL BE SAME SO

Q3= 7.5×10-4 C, V3=Q3/C3=7.5×10-4/15×10-6 = 50V

U=1/2×Q3×V3=187.5×10 -4 J

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