Turn in scantron You keep these question sheets

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1 Exam 2 on OCT Physics 106 R. Schad YOUR NAME ¼À Turn in scantron You keep these question sheets

2 1) This is to identify the exam version you have IMPORTANT Mark the A 2) This is to identify the exam version you have IMPORTANT Mark the B 3) Originally the capacitor was empty. Once we close the switch, the currents through r 1, r 2 and the capacitor will evolve like: [r 1 = 1 k; r 2 = 3 k; C = 10 F;] a) The current in r1 increases with time and The current in r2 increases with time and The current in C decreases with time b) The current in r1 decreases with time and The current in r2 increases with time and The current in C increases with time c) The current in r1 increases with time and The current in r2 decreases with time and The current in C decreases with time d) The current in r1 decreases with time and The current in r2 decreases with time and The current in C decreases with time e) The current in r1 decreases with time and The current in r2 increases with time and The current in C decreases with time

3 4) R R 2 Charges: one positive of unknown magnitude and one negative of -1 C are a distance R apart. At the point marked x, a distance R to the right of Q, the net electric potential is zero. Then the electric potential at the midpoint between the 2 charges must be: A. Positive B. Negative C. Zero 5) Discharging a capacitor through a resistor, after 4 seconds ½ of the original charge is lost. At which time, measured from the start, are ¾ of the original charge lost? A) 4 s B) 6 s C) 8 s D) 12 s E) Something else 6) A negative charge is shot into an electric field in the +x direction and slowing down. From this we can conclude that the electric potential a) Is higher on the left than on the right. b) Is higher on the right than on the left. c) Is constant d) none of these

4 7) Two +1 charges are located on the x-axis at x = -1 and x = 1 respectively. Rank the magnitude of the electric potential at the following points A,B,C,D given as [x-value,y-value] A = [0, 0] B = [0, 1] C = [-2, 0] D = [2, 0] A) A > B > C =D B) C = D > A > B C) A = B > C = D D) A > B = C = D E) Something else 8) A parallel plate capacitor of capacitance C 0 has plates of area A with separation d between them. When it is connected to a battery of voltage V 0, it has charge of magnitude Q 0 on its plates and energy E0 stored in it. It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them. After the plates are 2d apart, the magnitude of the charge on the plates and the energy stored in the capacitor are: a. 1 1 Q0, E0 2 2 b. Q 0, 2 1 E0 c. Q 0, E 0 d. Q 0, 2E 0 e. 2Q 0, 2E 0

5 9) The capacitors (2F and 1F) were connected in series to and got charged by a battery. If the energy stored on the 1 F capacitor is U 1F = 16 J, the energy on the 2 F capacitor must be: a. U 2F = 16 J b. U 2F = 8 J c. U 2F = 4 J d. U 2F = 32 J e. U 2F = 256 J ) You have two identical capacitors. They can be connected in series or in parallel or you can use just one of them. If you want the largest capacitance for the combination, how do you use them? a) 2 in series b) 2 in parallel c) use only one of them d) either way; all combinations have the same capacitance 11) In the circuit, the energy stored in C 1 U C1 = 16 J. C 1 = C 2 = C 3 = 1 F then the energy stored in C 3 is a) U 3 = 4 J b) U 3 = 8 J c) U 3 = 16 J d) U 3 = 32 J e) U 3 = 64 J 12) R1 = 4 R2 = 2 R3 = 1 V Bat = 40 V How does the electric power dissipated in each of the 3 resistors compare? Bat. + - R1 R2 R3 a) P 1 = 2P 2 = 4P 3 b) P 1 = 4P 2 = 16P 3 c) 4P 1 = 2P 2 = P 3 d) 16P 1 = 4P 2 = P 3 e) P 1 = P 2 = P 3

6 13) The capacitors [all the same capacitance] got connetcted as shown to the battery with voltage V and charged up. If the voltage drop across C3 came out to be 10 V, Then the voltage drop across C1 must be: a) 40 V b) 20 V c) 10 V d) 5 V e) 2.5 V 14) In the circuit, what is the battery voltage V Bat if the charge on C 3 = 1 C? [capacitance: C 1 = 1 F; C 2 = 1 F; C 3 = 2 F] a) V Bat = zero b) V Bat = 0.5 V c) V Bat = 1.0 V d) V Bat = 2.0 V e) V Bat = 3.0 V 15) When the current through the far-right 5 resistor equals I 5 = 5 A, the current through the lower-left 5 resistor then is: a. 5 A b. 20 A c. 15 A d A e. None of these + 5 Ω 5 Ω 10 Ω 5 Ω

7 16) The circuit to the right contains the usual suspects. V = 15 V R1 = 2 R2 = 3 C = 10 F Before we start, the switch is open, no charge on the capacitor. Then we close the switch. In the very first moment, the voltage V2 across R2 is: a) 15 V b) 10 V c) 9 V d) 7.5 V e) zero 17) R1 = 4 R2 = 2 R3 = 1 V Bat = 40 V How does the current flowing through each of 3 resistors compare? a) I 1 = 2I 2 = 4I 3 b) I 1 = 4I 2 = 16I 3 c) 4I 1 = 2I 2 = I 3 d) 16I 1 = 4I 2 = I 3 e) I 1 = I 2 = I 3 Bat. + - R1 R2 R3

8 Kinematics Newton s Law v = v 0 + a t x = x 0 + v 0 t + ½ a t 2 v 2 = v a (x x 0 ) v = (v + v 0 ) / 2 F = m a F gravity = m g g = 9.80 m/s 2 Conservation of Energy KE 1 + U 1 + W in/out = KE 2 + U 2 Energy Kinetik (linear): KE lin = ½ mv 2 Potential (gravity): U g = m g y Work Power (electrical) W = F d = F d cos P = W/t = E/t P = I V = I 2 R = (V) 2 / R Coulomb force F = k e q 1 q 2 / r 2 along the connecting line Electric field Electric flux Gauss Law Potential energy Potential E = F/q = E = k e q / r 2 dq r ke 2 volume E gradient E = E = surface E d A closedsurface U = U B - U A = V = U / q = V = k e q/r V for a point charge / pointing radially dv dx E d A = q inside / o = 4k e q inside B q E d s = q V A B A E d s for a point charge

9 Capacitance C = Q/V [ = o A/d parallel plate C] C eq = C 1 + C 2 + C 3 + [parallel combination] 1/C eq = 1/C 1 + 1/C 2 + 1/C 3 + [series combination] U = ½ Q 2 /C = ½ Q V = ½ C (V) 2 [energy stored in C] Charging/discharging of Capacitor q(t) = Q (1 - e -t/rc ) I(t) = (V/R) e -t/rc q(t) = Q e -t/rc I(t) = -(V/R) e -t/rc Ohm's Law R = V/I = l/a [ = resistivity = 1/] Resistivity / Resistance electron mass proton mass elementary charge Coulomb constant Permittivity of free space = m e / (n q 2 ) [ = scattering time] = o [1 + (T T o )] [temperature dependence] 1/R eq = 1/R 1 + 1/R 2 + 1/R 3 + [parallel] R eq = R 1 + R 2 + R 3 + [series] me = kg mp = kg e = C ke = Nm 2 /C 2 o = C 2 /Nm 2 [electron: -e ; proton: +e] [ke = 1/4o]