midterm practice exam 2 solution

# Physics for Scientists and Engineers (3rd Edition)

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PHY132 Midterm-II Practice Exam Problems: 1) A R = 3 M resistor is connected across a C = 1 µ F capacitor that is charged to a voltage V 0 = 10 V. a) Write an expression for the voltage V on the capacitor as function of time. RC / t 0 RC / t 0 e V C ) t ( Q ) t ( V Ce V ) t ( Q = = = b) What is the voltage on the capacitor at t = 3s? V 7 . 3 e V ) s 3 ( V s 3 V As 10 x A V 10 x 3 RC 1 0 6 6 = = = = c) What is the current through the resistor at t = 3s? A 2 . 1 e A / V 10 x 3 V 10 ) s 3 t ( I e R V dt dQ I 1 6 RC / t 0 µ = = = = = d) What is the total amount of energy dissipated in the resistor during the discharging of the capacitor? All the energy that initially is stored in the electric field of the capacitor is dissipated during this process (where else should it go?). Therefore: J 10 x 5 ) V 10 ( x V C 10 x 2 1 CV 2 1 U 5 2 6 2 = = = 2) Four currents I 1 , I 2, I 3, and I 4 are send through the system of wires as shown in the diagram. The straight wires point directly to the center of the circle formed by the curved parts of the wires. The circle has a radius r = 10 cm and all currents have a magnitude of 4 A. a) What is the magnetic field at the center of the circle generated by one of the straight wires? A B C D Since the straight wires are parallel to the lines from the center of the circle to the wires, the Biot-Savart Law tells you that the magnetic field at the center of the circle generated by each one of the straight wires is ZERO.

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b) What is the magnetic field at the center of the circle generated by the curved segment A? In which direction does it point? Use the Biot-Savart Law:
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## This homework help was uploaded on 02/04/2008 for the course PHY 132 taught by Professor Rijssenbeek during the Fall '04 term at SUNY Stony Brook.

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midterm practice exam 2 solution - PHY132 Midterm-II...

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