0306phk - 1 - July 17, 20031. The potential difference across the 10 W - PDF document

0306phk - 1 - July 17, 20031. The potential difference across the 10 W resistor is 7.1 V. = 15.0 V a)What is the power dissipated by the VIR07107181576576|||| 0308phk - 2 - September 4, 20032. a) For the circuit below, what is the terminal voltage of the battery? (4 marks) 12.0 V ====-=¨12013313340531205367...dropVb)If resistor terminal voltage?(1 mark) R
increase
no change
 decrease c)Using principles of physics, explain your choice for b). Additional R in parallel results in an overall lower R, thus an increase in current.(2 marks) As a consequence, a greater voltage drop Ir occurs across the internalresistance resulting in a smaller terminal voltage. (2 marks) 0106phk - 4 - July 25, 20013. The current through the 5  resistor in the circuit below is 0.14 A. 0.8010.033.0 50.068.0 a)Determine the emf of the battery. 0.1450.07.0V1mark 7.033.0 0.21A markI4V||R4 7.068.0 0.10A 0.210.140.100.45A1mark 1R3 1R4 15.4 10.015.40.8026.2 0.4526.211.9V12V1mark 0.40W 10.015.00.40W12.0V What is the resistance of R2?(7Êmarks) Key:P=I2RP1=I2R1I=P1R1 æèö12 0.40 æèö12 0.20A0.2100.2152marks1mark1mark1mark1mark1mark 5.A 12 V battery from a car is used to operate a 65 W headlight. a)How much energy does the headlight use in 1.5 hours? EPt12 1.536001mark3.5 b)What total charge passes through the headlight during this time? QEV 12 markQIt12 3.512V 2marksOR5.42A5400s2marks29000C =29000C c)What is the total number of electrons that pass through the headlight during this time period?(2 marks) NQe 1 mark290001.6 1 mark1.8 6.A 12 V battery transfers 33 C of charge to an external circuit in 7.5 s. a)What current flows through the circuit? \n \f )\r\t \f \t*1markb)What is the resistance of the circuit? \f\r \f \t \f \b\t\r1markc)What is the power output of the battery?

\r\r\f \b\t 1markd)The external circuit is most likely to consist of  a bulb. a kettle. a calculator. (Check one response.)(1 mark) 0201phk - 8 - March 4, 2002 7. What is the potential difference across the 6 . 0 W resistor in the circuit shown? (7 marks) 15. 0 W 7. 0 W 6. 0 W 5. 0 W 9. 0 W V = 8. 0 V R p 1 = 15 . 0 W + 6. 0  9. 0 W = 3 . 0 W 1 mark 1 R p = 1 7. 0 + 1 30 . 0 R p = 5 6 8 W  1 mark R T = 5. 0 + 5. 6 8 = 10 . 6 8 W  1 mark I T = V T R T = 8. 0 10 . 6 8 = 0. 7 5 A  1 mark V p = V T - V 5 = 8. 0 V - 0. 7 5 ´ 5. 0 = 4. 2 5 V  1 mark I p = V p R p = 4. 2 5 30 . 0 = 0 . 142 A  1 mark V 6 = I p R = 0 . 142 ´ 6. 0 = 0 . 85 V  1 mark 001phk- 9 -February 24, 200088 battery and two light bulbs. Each light bulb dissipates 5.0W 1.50A8.00V a)If a current of 1.50A (4Êmarks) Resistance Solution:Voltage Solution:Power Solution: 5.01.50 2.221mark 8.001.50 5.331mark5.3322.221mark0.891mark1.5V3.3V3.36.76.71.50.891mark1mark1mark1mark1.5812W1mark1mark1mark1.5 0.891mark 001phk- 10 -February 24, 2000b 8.00V S Lamp A will now be(1Êmark) i) p 4 brighter. pthe same brightness as before. pdimmer. (Check one response.) The batteryÕs terminal voltage will now be(1Êmark) ii)pgreater than before. pthe same as before. p 4 less than before. (Check one response.) c (3Êmarks) 1mark1mark1mark 0301phk- 9 -February 25, 20039. The terminal voltage of the battery is 5.8 V. 8.01611r=1.0S a)What is the emf of this battery? 65801451450405804010VIr...b)What is the effect on the emf of the battery when switch S is opened? no effect 0208phk- 9 -August 29, 200210. How much energy does the 6 2 resistor dissipate in 15 seconds in the circuit shown? (7 marks) 12.0V15.09.06.05.0 1Rp 9.0 6.0 3.615.03.65.023.6 12.023.6 508A12.00.5115.00.515.01.83 1.836.0 305A1.838.4J 0206phk- 8 -July 23, 200211. Each of the two cells shown is connected to an external 6 00. resistor. 1.5V0.500.25 1.5V \b 6.00 With supporting calculations, state which cell delivers the greater power to the 600 (7 marks) 6.00 1mark1.56.50 0.231mark0.236.000.32W1mark6.00 1mark1.56.25 0.241mark0.246.000.35W1mark1mark