Question

18] #. Match the name below with the device at the right that it represents. I. Et Il Resistance. ideal emf (8) device. Capacitance Real battery IV. [12] #2. Circle all the TRUE statements below. a) The Loop Rule is simply a statement of Conservation of Energy in an electrical circuit. b) The Junction Rule is simply a statement of Conservation of Electric Charge in an electrical circuit. e) When measuring the electric potential difference across a component in an e parallel with the component d) When measuring the electric current in a component in an electrical circuit, the meter must be connected in series component e) The resistivity pm) of a piece of wire depends on its length D If a piece of copper wire (a metal) is cooled to a low temperature then the resistivity of it will increase. lectrical circuit, the meter must be com [21] #3. For the circuit shown below, write three loop rule equations and a useful junction rule equation involving f 5.0V 152 8 2 i 162 Loop Loop Junction +-3.0 V 9.0 V [4] #4. The capacitance of an air-filled capacitor is 3600 pF capacitor. Determine the new capacitance value when a whose dielectric constant is 7.9 is placed to completely fill the space between the plates. Answer: (a) 28.4 nF. (b) 30.8 nF. ()36.3 nF. (d)41.2 nF. (e) 442 nF. [8] #5. For the circuit shown, R1-220 ka, R2-350 ka R 420 k2 and R-590 k2. Determine the equivalent between a and b. R: Rs Answer: (a) 0.27 MS2. (b) 0.35 Min (d) 0.54 Mia (e) 65 MQ. (f) (c) 0.46 Mn- A R4 [10] #6. Circle all the FALSE statements below. a) The vector equation F-qvB gives the force F on a particle with charge q that is moving with velocity The right-hand rules given apply to only negative charges. c) magnetic field B. b) Like magnetic poles repel and unlike magnetic poles attract. The Hall effect can be used to determine whether the "majority" charge carriers in a material have a positive, or, negative charge d) e) in P-type semiconductor materials, the majority charge carriers "looks" like electrons, except that they ar 8 #7. A heating element is made by maintaining a potential difference of 120 V between the ends of a certain win 7.52 x 10- m2 cross section area and a resistivity of 4.86 x 106 ohm.m. If the element dissipates 10.0 W, what is its Answer: (a) 0.01 m. (b) 0.08 m. (c) 12.4 m. (d) 18.4 m. (e) 22.3 m. (0) Problems #8, 9 and 10 refer to the same wire. [6] #8. An electrical current of 158 mA exists in a solid cylindrical wire whose diameter is 3.50 mm and length is 10 Calculate the magnitude of the current density in the wire. Answer: (a) 10.8 kA/m2. (b) 13.4 kA/m2. (e) 16.4 kA/m2. (d) 18.6 kA/m2. (e)20.4 kA/m2. (

Answer 1