BROCK UNIVERSITY Physics 1P22/1P92 Mid-term Test 2: 19 March 2010 Solutions 1. [6 marks] (See Page 746, CP # 24, and pages 15 16 of the posted Ch. 22 lecture notes from 4 March.) 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 1.5 10 6 Ω m. Calculate the current in the wire. Solution: Strategy: First calculate the resistance of the wire, then use Ohm s law to determine the current. Step 1: R = ρl A = ρl πr 2 = 1.5 10 6 (0.5) π(4 10 3 ) 2 = 1.492078 Ω Using Ohm s law, the current through the wire is I = V R 3 = 1.492078 = 2.01 A Thus, the current through the wire is 2.0 A.
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. S olution: This is very similar to Example 21.12 on Page 709 of the textbook. I ll solve it in a slightly different way than shown in the textbook (but check the textbook method to see which you prefer). The capacitance C 1 before the dielectric is inserted is increased to C 2 = κc 1 after the dielectric is inserted. Noting that Q = C V, and also noting that the charge on the plates is constant (because the battery is disconnected before the dielectric is inserted), we conclude that C 1 V 1 = C 2 V 2 Therefore, C 2 = V 1 C 1 V 2 But the ratio in the previous equation is the dielectric constant κ, so κ = C 2 C 1 = V 1 V 2 = 10 4.0 = 2.5 Thus, the dielectric constant of the dielectric is 2.5.
3. [6 marks] Calculate the equivalent resistance for the entire circuit. Solution: (This is part of CP # 60 on Page 785; see pages 17ff of the Ch 23 lecture notes.) The 8 Ω resistor is in parallel with the bottom 24 Ω resistor; their equivalent resistance is ( 1 R 1 = 8 + 1 ) 1 ( 3 = 24 24 + 1 ) 1 ( ) 1 4 = = 24 24 24 4 = 6 Ω This equivalent resistance R 1 is in series with the 6 Ω resistor, so their equivalent resistance is R 2 = R 1 + 6 = 6 + 6 = 12 Ω Equivalent resistance R 2 is in parallel with the 24 Ω resistance to the right, so the equivalent resistance of this combination is R 3 = ( 1 12 + 1 ) 1 ( 2 = 24 24 + 1 ) 1 ( ) 1 3 = = 24 24 24 3 = 8 Ω Equivalent resistance R 3 is in series with the 4 Ω resistance, so the final equivalent resistance of the entire circuit is R = R 3 + 4 = 8 + 4 = 12 Ω
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. Solution: (This is somewhat like Page 780, CQ # 18.) First calculate the total equivalent resistance of the circuit, then calculate the total current flowing from the battery, then calculate the individual currents through each light bulb, and finally determine the power dissipated in each bulb. The equivalent resistance of the two light bulbs in parallel is ( 1 R 1 = 10 + 1 ) 1 = 5 Ω 10 Thus, the total equivalent resistance of the circuit is 10 + 5 = 15 Ω. This means that the total current from the battery is I = V R = 10 15 = 0.6667 A All of the current from the battery passes through light bulb A, so I A = 0.67 A. Therefore, the power dissipated by light bulb A is P A = I 2 AR = 0.6667 2 (10) = 4.4 W The voltage across light bulb A is V = I A R = 6.7 V. Therefore, the voltage across each of light bulbs B and C is 10 6.7 = 3.3 V. This means that the current through each of light bulbs B and C is I B = I C = V R = 3.3 10 = 0.33 A The power dissipated by each of the light bulbs B and C is P B = P C = I 2 BR = 1.1 W
5. [4 marks] Consider the same circuit as in Question 4 (the figure is reproduced here for your convenience). Suppose that an ideal wire is added to the circuit so that it connects points 1 and 2. Calculate the current flowing through each light bulb and through the new wire. Solution: All of the current from the battery flows through light bulb A and then through the additional wire that connects points 1 and 2. No current flows through light bulbs B and C, since they are both in parallel with the short-circuit wire that connects points 1 and 2. Thus, the equivalent resistance of the circuit is just the resistance of light bulb A, which is 10 Ω. Thus the current through light bulb A and the additional wire connecting points 1 and 2 is I = V R = 10 10 = 1 A The current through light bulbs B and C is 0 A.
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. TRUE FALSE EXPLANATION: There is only one pathway, so each electron passes through each resistance. As each electron passes through each resistance, its electric potential energy decreases, according to Ohm s law. (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. EXPLANATION: Current is not used up in a circuit; what is used up is electric potential energy (which is converted to thermal energy and perhaps light). The same current passes through each resistance in a series circuit; since each light bulb has the same resistance, each also has the same brightness. (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. EXPLANATION: Adding more branches to a parallel circuit decreases the equivalent resistance of the circuit, therefore increasing the overall current flowing from the battery. (d) The light bulbs in the following circuit all have the same resistance, and therefore they all have the same brightness. EXPLANATION: This is a parallel circuit with two branches. The branch with bulbs B and C has greater resistance than the branch with bulb A, so more current flows through bulb A. Since P = I 2 R, bulb A is brighter than bulbs B and C. (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. TRUE FALSE
EXPLANATION: Since C = ɛ 0 A/d, when the plates are farther apart, the capacitance 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. EXPLANATION: The charge remains constant, but the capacitance decreases (as we learned in Part (e)). This means that the voltage must increase, because Q = C V. (Remember that voltage is a measure of electric potential energy per unit charge; the work done in moving the plates increases the potential energy, just as the gravitational potential energy would increase if you lifted a mass to a higher level in a gravitational field.)