50 w then the network dissipates 200 w other

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dissipates 0.50 W, then the network dissipates 2.00 W. Other combinations also give the desired resistance and power rating, such as two in series in parallel with two in series. Figure 19.82 19.83. Set Up: Since they are in parallel, the voltage across each device is 120 V. The total current drawn from the outlet is the sum of the individual currents. Solve: (a) Toaster: Frypan: Lamp: (b) This is greater than 20 A and the circuit breaker will blow. 19.84. Set Up: The resistance of each half of a wire is The combination can be represented by the resistor network shown in Figure 19.84. Figure 19.84 Solve: The two resistors in parallel have an equivalent resistance of Then the three in series have an equiv- alent resistance of 19.85. Set Up: Let R be the resistance of each resistor. Solve: When the resistors are in series, and When the resistors are in parallel, Reflect: In parallel, the voltage across each resistor is the full applied voltage V. In series, the voltage across each resistor is and each resistor dissipates less power. 19.86. Set Up : For resistors in series, the currents are the same and the voltages add. For resistors in parallel, the currents add and the voltages are the same. 1 R eq 5 1 R 1 1 1 R 2 1 c R eq 5 R 1 1 R 2 1 c V / 3 P p 5 V 2 R / 3 5 3 V 2 R 5 9 P s 5 9 1 27 W 2 5 243 W. R eq 5 R / 3. P s 5 V 2 3 R . R eq 5 3 R P tot 5 V 2 R eq . 0.500 V 1 0.250 V 1 0.500 V 5 1.25 V . 0.250 V . 0.500 V 0.500 V 0.500 V 0.500 V 0.500 V . I 5 15.0 A 1 11.7 A 1 0.625 A 5 27.3 A. I 5 75 W 120 V 5 0.625 A. I 5 1400 W 120 V 5 11.7 A. I 5 P V 5 1800 W 120 V 5 15.0 A. P 5 VI . 1000 V 1000 V 1000 V 1000 V 1000 V . 500 V R eq 5 1000 V . Current, Resistance, and Direct-Current Circuits 19-21
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Solve: (a) Replacing series and parallel combinations of resistors by their equivalents gives the equivalent networks as shown in Figure 19.86. The equivalent resistance of the network is Figure 19.86 (b) The voltage across the resistor in Figure 19.86a is This is the voltage across the resistor in Figure 19.86b. The current through the resistor is 4.8 A. This is also the current through the resistor and the voltage across that resistor is 96.0 V. Therefore, the voltage between points x and y is This voltage is divided equally between the resistors in Figure 19.86b, so the voltage between points x and y is The voltmeter will read 57.6 V. 19.87. Set Up: Table 19.1 gives the resistivities of copper and aluminum to be and For the cables in series (end-to-end), For the cables in parallel the equivalent resistance is given by Note that in the two configurations the copper and aluminum sections have different lengths. And, for the parallel cables the cross sectional area of each cable is half what it is for the end to end configuration. Solve: end-to-end: for each cable. parallel: Now for each cable. L is doubled and A is halved compared to the other configuration, so and and The least resistance is for the cables in parallel. R eq 5 0.416 V . 1 R eq 5 1 R c 1 1 R a 5 1 0.688 V 1 1 1.052 V R a 5 4 1 0.263 V 2 5 1.052 V . R c 5 4 1 0.172 V 2 5 0.688 V L 5 1.00 3 10 3 m R eq 5 0.172 V 1 0.263 V 5 0.435 V R a 5 r a L A 5 1 2.63 3 10 2 8 V # m 21 0.50 3 10 3 m 2 0.500 3 10 2 4 m 2 5 0.263 V R c 5 r c L A 5 1 1.72 3 10 2 8 V # m 21 0.50 3 10 3 m 2 0.500 3 10 2 4 m 2 5 0.172 V L 5 0.50 3 10 3 m 1 R eq 5 1 R c 1 1 R a . R eq R eq 5 R c 1
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