Physics Solution Manual for 1100 and 2101

The right end of the resistor is and the left end is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) and the series combination (4.0 Ω and the 6.0 Ω) are in parallel; therefore, their equivalent resistance is 2.98 Ω. The 2.98-Ω combination is in series with the 3.0-Ω resistor, so that equivalent resistance is 5.98 Ω. Finally, the 5.98-Ω ELECTRIC CIRCUITS 1094 combination and the 20.0-Ω resistor are in parallel, so the equivalent resistance between the points A and B is 4 .6 Ω . ______________________________________________________________________________ 66. REASONING The two resistors R1 and R2 are wired in series, so we can determine their equivalent resistance R12. The resistor R3 is wired in parallel with the equivalent resistance R12, so the equivalent resistance R123 can be found. Finally, the resistor R4 is wired in series with the equivalent resistance R123. With these observations, we can evaluate the equivalent resistance between the points A and B. R1 = 16 Ω R2 = 8 R4 = 26 Ω A B R3 = 48 Ω SOLUTION Since R1 and R2 are wired in series, the equivalent resistance R12 is R12 = R1 + R2 = 16 Ω + 8 Ω = 24 Ω (20.16) The resistor R3 is wired in parallel with the equivalent resistor R12, so the equivalent resistance R123 of this combination is 1 1 1 1 1 = + = + R123 R3 R12 48 Ω 24 Ω or R123 = 16 Ω (20.17) The resistance R4 is in series with the equivalent resistance R123, so the equivalent resistance RAB between the points A and B is RAB = R4 + R123 = 26 Ω + 16 Ω = 42 Ω ______________________________________________________________________________ Chapter 20 Problems 1095 60.0 Ω 67. REASONING The circuit diagram is shown at the right. We can find the current in the 120.0-Ω resistor by 20.0 Ω A B using Ohm’s law, provided that we can obtain a value for VAB, the voltage between points A and B in the 120.0 Ω diagram. To find VAB, we will also apply Ohm’s law, this time by multiplying the current from the battery times RAB, the equivalent parallel resistance between A 15.0 V and B. The current from the battery can be obtained by applying Ohm’s law again, now to the entire circuit and using the total equivalent resistance of the series combination of the 20.0-Ω resistor and RAB. Once the current in the 120.0-Ω 2 resistor is found, the power delivered to it can be obtained from Equation 20.6b, P = I R. SOLUTION a. According to Ohm’s law, the current in the 120.0-Ω resistor is I120 = VAB/(120.0 Ω). To find VAB, we note that the equivalent parallel resistance between points A and B can be obtained from Equation 20.17 as follows: 1 1 1 = + RAB 60.0 Ω 120.0 Ω or RAB = 40.0 Ω This resistance of 40.0 Ω is in series with the 20.0-Ω resistance, so that according to Equation 20.16, the total equivalent resistance connected across the battery is 40.0 Ω + 20.0 Ω = 60.0 Ω. Applying Ohm’s law to the entire circuit, we can see that the current from the battery is 15.0 V I= = 0.250 A 60.0 Ω Again applying Ohm’s law, this time to the resistance RAB, we find that VAB = ( 0.250 A ) RAB = ( 0.250 A ) ( 40.0 Ω ) = 10.0 V Finally, we can see that the current in the 120.0-Ω resistor is I120 = VAB 10.0 V –2 = = 8.33 × 10 A 120 Ω 120 Ω b. The power delivered to the 120.0-Ω resistor is given by Equation 20.6b as ( P = I120 R = 8.33 × 10 A 2 –2 ) 2 (120.0 Ω ) = 0.833 W ______________________________________________________________________________ 1096 68. ELECTRIC CIRCUITS REASONING The total power P delivered by the battery is related to the equivalent resistance Req connected between the battery terminals and to the battery voltage V according to Equation 20.6c: P = V 2 / Req . We note that the combination of resistors in circuit A is also present in circuits B and C (see the shaded part of these circuits in the following drawings). In circuit B an additional resistor is in parallel with the combination from A. The equivalent resistance of resistances in parallel is always less than any of the individual resistances alone. Therefore, the equivalent resistance of circuit B is less than that of A. In circuit C an additional resistor is in series with the combination from A. The equivalent resistance of resistances in series is always greater than any of the individual resistances alone. Therefore, the equivalent resistance of circuit C is greater than that of A. We conclude then that the equivalent resistances are ranked C, A, B, with C the greatest and B the smallest. R R R R R R R R R +− V B +− V A R R R R R +− V C Since the total power delivered by the battery is P = V 2 / Req , it is inversely proportional to the equivalent resistance. The battery voltage V is the same in all three cases, so the power ranking is the reverse of the ranking deduced previously for Req. In other words, we expect that, from greatest to smallest, the total power delivered by the battery is B, A, C. SOLUTION The total power delivered by the battery is P = V 2 / Req . The voltage is given, but we must determine the equivalent resistance in each...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online