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1 Name: Student #: BROCK UNIVERSITY Page 1 of 8 Mid-term Test 2: March 2010 Number of pages: 8 Course: PHYS 1P22/1P92 Number of students: 125 Examination date: 19 March 2010 Number of hours: 2 Time of Examination: 19:30 21:30 Instructor: S. D Agostino A formula sheet is attached to the test paper. No other aids are permitted except for a non-programmable, non-graphing calculator. Solve all problems in the space provided. Total number of marks: [6 marks] A 3.0 V potential difference is applied between the ends of a nichrome wire of diameter 0.80 mm and length 50 cm. The resistivity of nichrome is Ω m. Calculate the current in the wire.

2 PHYS 1P22/1P92 March 2010 Page 2 of 8 2. [6 marks] A parallel-plate capacitor is charged using a 10 V battery; then the battery is removed. A dielectric slab is then slid between the plates of the capacitor, filling all the space between the plates; this reduces the voltage difference between the plates to 4.0 V. Calculate the dielectric constant of the dielectric.

3 PHYS 1P22/1P92 March 2010 Page 3 of 8 3. [6 marks] Calculate the equivalent resistance for the entire circuit.

4 PHYS 1P22/1P92 March 2010 Page 4 of 8 4. [6 marks] Consider the figure, and suppose the battery s voltage is 10 V and each light bulb has resistance 10 Ω. (a) Calculate the current flowing through each light bulb. (b) Calculate the power dissipated by each light bulb.

5 PHYS 1P22/1P92 March 2010 Page 5 of 8 5. [4 marks] Consider the same circuit as in Question 4 (the figure is reproduced here for your convenience); the battery s voltage is still 10 V and each light bulb still has resistance 10 Ω. Suppose that an ideal wire is added to the circuit (not shown in the figure, but draw it in yourself if you wish) so that it connects points 1 and 2. Calculate the current flowing through each light bulb and through the new wire.

6 PHYS 1P22/1P92 March 2010 Page 6 of 8 6. [12 marks] Indicate whether each statement is TRUE or FALSE. Then provide a BRIEF explanation (i.e. one or two sentences). Your explanation may include formulas, if you wish. Remember, brevity and clarity are courtesy. (a) In a series circuit with one battery and several resistances, the electric potential energy of each electron decreases as it moves through the circuit outside the battery. (b) In a series circuit that contains a battery and two light-bulbs with equal resistance, the bulb closer to the negative side of the battery is brighter. This happens because some current is used up in the first light-bulb, so there is not as much left by the time it reaches the second light bulb. (c) When additional parallel branches are added to a parallel circuit, where each new branch has some resistance, the overall resistance of the circuit increases, and therefore the overall current flowing from the battery decreases.

7 PHYS 1P22/1P92 March 2010 Page 7 of 8 (d) The light bulbs in the following circuit all have the same resistance, and therefore they all have the same brightness. (e) A parallel-plate capacitor is fully charged by a 10 V battery, after which the battery is disconnected. The plates are then moved further apart using insulating handles. After the plates are moved, the capacitance of the capacitor decreases. (f) A parallel-plate capacitor is fully charged by a 10 V battery, after which the battery is disconnected. The plates are then moved further apart using insulating handles. After the plates are moved, the voltage difference between the plates is still 10 V.

8 PHYS 1P22/1P92 March 2010 Page 8 of 8 Question Total Marks Out of Potentially useful formulas F = K q 1 q 2 r 2 U elec = qv E = F q E = V d Q = C V parallel-plate capacitor: E = Q ɛ 0 A C = κɛ 0A d point charge: E = K q r 2 point charge: V = K q r E = E κ and U C = 1 2 Q2 C = 1 2 C( V )2 u E = U C volume = 1 2 κɛ 0E 2 C = ɛ 0A d R = ρl A I = Q t V = IR P = U t = I V = I2 R = ( V )2 R Kirchhoff: Iin = I out V loop = V i = 0 i R eq = R 1 + R 2 + R R N R eq = ( ) 1 R 1 R 2 R 3 R N K = N m2 C 2 ɛ 0 = C 2 N m 2 mass of proton (and neutron) = kg mass of electron = kg fundamental unit of charge: e = C

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