26 - DIRECT-CURRENT CIRCUITS 26 26.7.IDENTIFY: First do as...

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D IRECT -C URRENT C IRCUITS 26.7.IDENTIFY: First do as much series-parallel reduction as possible. SET UP: The 45.0-Ω and 15.0-Ω resistors are in parallel, so first reduce them to a single equivalent resistance. Then find the equivalent series resistance of the circuit. EXECUTE: 1/ R p = 1/(45.0 Ω) + 1/(15.0 Ω) and R p = 11.25 Ω. The total equivalent resistance is 18.0 Ω + 11.25 Ω + 3.26 Ω = 32.5 Ω. Ohm’s law gives I = (25.0 V)/(32.5 Ω) = 0.769 A. EVALUATE: The circuit appears complicated until we realize that the 45.0-Ω and 15.0-Ω resistors are in parallel. 26.8. IDENTIFY: Eq.(26.2) gives the equivalent resistance of the three resistors in parallel. For resistors in parallel, the voltages are the same and the currents add. (a) SET UP: The circuit is sketched in Figure 26.8a. EXECUTE: parallel eq 1 2 3 1 1 1 1 R R R R eq 1 1 1 1 1.60 2.40 4.80 R eq 0.800 R Figure 26.8a (b) For resistors in parallel the voltage is the same across each and equal to the applied voltage; 1 2 3 28.0 V V V V E 1 1 1 28.0 V so 17.5 A 1.60 V V IR I R 23 28.0 V 28.0 V 11.7 A and 5.8 A 2.40 4.8 VV II RR (c) The currents through the resistors add to give the current through the battery: 1 2 3 17.5 A 11.7 A 5.8 A 35.0 A I I I I EVALUATE: Alternatively, we can use the equivalent resistance eq R as shown in Figure 26.8b. eq 0 IR E eq 28.0 V 35.0 A, 0.800 I R E which checks Figure 26.8b (d) As shown in part (b), the voltage across each resistor is 28.0 V. (e) IDENTIFY and SET UP: We can use any of the three expressions for 22 : / . P P VI I R V R They will all give the same results, if we keep enough significant figures in intermediate calculations. 26
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EXECUTE: Using 2 /, P V R 22 1 1 1 2 2 2 28.0 V 28.0 V / 490 W, / 327 W, and 1.60 2.40 P V R P V R 2 2 3 3 3 28.0 V / 163 W 4.80 P V R EVALUATE: The total power dissipated is out 1 2 3 980 W. P P P P This is the same as the power in 2.80 V 35.0 A 980 W PI E delivered by the battery. (f) 2 /. P V R The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance. 26.10. (a) IDENTIFY: The current, and hence the power, depends on the potential difference across the resistor. SET UP: / P V R EXECUTE: (a) (5.0 W)(15,000 ) 274 V V PR (b) / (120 V) /(9,000 ) 1.6 W P V R SET UP: (c) If the larger resistor generates 2.00 W, the smaller one will generate less and hence will be safe. Therefore the maximum power in the larger resistor must be 2.00 W. Use 2 P I R to find the maximum current through the series combination and use Ohm’s law to find the potential difference across the combination. EXECUTE: P I R gives I = P / R = (2.00 W)/(150 Ω) = 0.0133 A. The same current flows through both resistors, and their equivalent resistance is 250 Ω. Ohm’s law gives V = IR = (0.0133 A)(250 Ω) = 3.33 V. Therefore P 150 = 2.00 W and 100 P I R = (0.0133 A) 2 (100 Ω) = 0.0177 W. EVALUATE: If the resistors in a series combination all have the same power rating, it is the largest resistance that limits the amount of current. 26.13. IDENTIFY: In both circuits, with and without R 4 , replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use 2 P I R to calculate the power dissipated in each bulb.
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This note was uploaded on 10/25/2010 for the course PHYS 260 taught by Professor Hkmiet during the Spring '08 term at George Mason.

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26 - DIRECT-CURRENT CIRCUITS 26 26.7.IDENTIFY: First do as...

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