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

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D IRECT -C URRENT C IRCUITS 26.7.I DENTIFY : First do as much series-parallel reduction as possible. S ET U P : The 45.0- Ω and 15.0 - Ω resistors are in parallel, so fi rst reduce them to a single equivalent resistance. Then find the equivalent series resistance of the circuit. E XECUTE : 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. E VALUATE : The circuit appears complicated until we realize that the 45.0- Ω and 15.0 - Ω resistors are in parallel. 26.8. I DENTIFY : 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) S ET U P : The circuit is sketched in Figure 26.8a. E XECUTE : 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 2 3 2 3 2 3 28.0 V 28.0 V 11.7 A and 5.8 A 2.40 4.8 V V I I R R (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 E VALUATE : 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) I DENTIFY and S ET U P : We can use any of the three expressions for 2 2 : / . 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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E XECUTE : Using 2 / , P V R 2 2 2 2 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 E VALUATE : 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 P I 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) I DENTIFY : The current, and hence the power, depends on the potential difference across the resistor. S ET U P : / P V R E XECUTE : (a) (5.0 W)(15,000 ) 274 V V PR (b) / (120 V) /(9,000 ) 1.6 W P V R S ET U P : (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. E XECUTE : P I R gives I = P / R = (2.00 W)/(150 Ω) = 0.0133 A. The same cu rrent 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. E VALUATE : If the resistors in a series combination all have the same power rating, it is the largest resistance that limits the amount of current.
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