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R1 = 3 N 9 V R2 = 70 R3 = 9 2 FIGURE 12.1 3 The circuit for Example 12.4. The light bulbs are connected in a series arrangement; therefore, the total resistance is given by Ruve = R, + R, + R, = 3 0 + 7 Q + 9 N = 19 Q We can now use Ohm's law to determine the current flowing through the circuit. V 0.47 A 2 0.5 A 19 R. total We can also obtain the voltage drop across each lamp using Ohm's law: V-2 = R,I = (3)(0.47) = 1.41 V V3 = R,I = (7)(0.47) = 3.29 V V34 = R,I = (9)(0.47)= 4.23 V 2-3 Note that, neglecting rounding-off errors, the sum of the voltage drops across each light bulb should be 9 volts. |R = 3 N R2 = 7 N R3 = 9 N 9 V FIGURE 12.1 5 The circuit for Example 12.5. Because the light bulbs are connected in a parallel arrangement, the voltage drop across each light bulb is equal to 9 volts. We use Ohm's law to determine the current in each branch in the following manner: V = R,1, = 9= 31, = I, = 3.0 A V = R,1, = 9=71, = 1,=1.3 A V = R1, = 9= 31, = 1,=1.0 A The total current drawn by the circuit is Ignal = 1, + 1, + I, = 3.0 + 1.3 + 1.0 = 5.3 A total The total resistance is given by 1 R,' R, 1 1 + 1 1 +=+ 7'9 1 R = 1.70 R, 3 total totat Note that we could have obtained the total current drawn by the circuit using the total resistance and the Ohm's law in the following manner: V = R - 9 V= (1.7 0)(Ia) = Ie = 5.3 A total* total total total

R1 = 3 N 9 V R2 = 70 R3 = 9 2 FIGURE 12.1 3 The circuit for Example 12.4. The light bulbs are connected in a series arrangement; therefore, the total resistance is given by Ruve = R, + R, + R, = 3 0 + 7 Q + 9 N = 19 Q We can now use Ohm's law to determine the current flowing through the circuit. V 0.47 A 2 0.5 A 19 R. total We can also obtain the voltage drop across each lamp using Ohm's law: V-2 = R,I = (3)(0.47) = 1.41 V V3 = R,I = (7)(0.47) = 3.29 V V34 = R,I = (9)(0.47)= 4.23 V 2-3 Note that, neglecting rounding-off errors, the sum of the voltage drops across each light bulb should be 9 volts. |R = 3 N R2 = 7 N R3 = 9 N 9 V FIGURE 12.1 5 The circuit for Example 12.5. Because the light bulbs are connected in a parallel arrangement, the voltage drop across each light bulb is equal to 9 volts. We use Ohm's law to determine the current in each branch in the following manner: V = R,1, = 9= 31, = I, = 3.0 A V = R,1, = 9=71, = 1,=1.3 A V = R1, = 9= 31, = 1,=1.0 A The total current drawn by the circuit is Ignal = 1, + 1, + I, = 3.0 + 1.3 + 1.0 = 5.3 A total The total resistance is given by 1 R,' R, 1 1 + 1 1 +=+ 7'9 1 R = 1.70 R, 3 total totat Note that we could have obtained the total current drawn by the circuit using the total resistance and the Ohm's law in the following manner: V = R - 9 V= (1.7 0)(Ia) = Ie = 5.3 A total* total total total

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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The light bulbs in the circuit as shown are placed in a parallel arrangement, as shown . Determine the current flow through each branch. Also compute the total resistance offered by all light bulbs to current flow.

R1 = 3 N 9 V R2 = 70 R3 = 9 2 FIGURE 12.1 3 The circuit for Example 12.4. The light bulbs are connected in a series arrangement; therefore, the total resistance is given by Ruve = R, + R, + R, = 3 0 + 7 Q + 9 N = 19 Q We can now use Ohm's law to determine the current flowing through the circuit. V 0.47 A 2 0.5 A 19 R. total We can also obtain the voltage drop across each lamp using Ohm's law: V-2 = R,I = (3)(0.47) = 1.41 V V3 = R,I = (7)(0.47) = 3.29 V V34 = R,I = (9)(0.47)= 4.23 V 2-3 Note that, neglecting rounding-off errors, the sum of the voltage drops across each light bulb should be 9 volts.
Transcribed Image Text:R1 = 3 N 9 V R2 = 70 R3 = 9 2 FIGURE 12.1 3 The circuit for Example 12.4. The light bulbs are connected in a series arrangement; therefore, the total resistance is given by Ruve = R, + R, + R, = 3 0 + 7 Q + 9 N = 19 Q We can now use Ohm's law to determine the current flowing through the circuit. V 0.47 A 2 0.5 A 19 R. total We can also obtain the voltage drop across each lamp using Ohm's law: V-2 = R,I = (3)(0.47) = 1.41 V V3 = R,I = (7)(0.47) = 3.29 V V34 = R,I = (9)(0.47)= 4.23 V 2-3 Note that, neglecting rounding-off errors, the sum of the voltage drops across each light bulb should be 9 volts.
|R = 3 N R2 = 7 N R3 = 9 N 9 V FIGURE 12.1 5 The circuit for Example 12.5. Because the light bulbs are connected in a parallel arrangement, the voltage drop across each light bulb is equal to 9 volts. We use Ohm's law to determine the current in each branch in the following manner: V = R,1, = 9= 31, = I, = 3.0 A V = R,1, = 9=71, = 1,=1.3 A V = R1, = 9= 31, = 1,=1.0 A The total current drawn by the circuit is Ignal = 1, + 1, + I, = 3.0 + 1.3 + 1.0 = 5.3 A total The total resistance is given by 1 R,' R, 1 1 + 1 1 +=+ 7'9 1 R = 1.70 R, 3 total totat Note that we could have obtained the total current drawn by the circuit using the total resistance and the Ohm's law in the following manner: V = R - 9 V= (1.7 0)(Ia) = Ie = 5.3 A total* total total total
Transcribed Image Text:|R = 3 N R2 = 7 N R3 = 9 N 9 V FIGURE 12.1 5 The circuit for Example 12.5. Because the light bulbs are connected in a parallel arrangement, the voltage drop across each light bulb is equal to 9 volts. We use Ohm's law to determine the current in each branch in the following manner: V = R,1, = 9= 31, = I, = 3.0 A V = R,1, = 9=71, = 1,=1.3 A V = R1, = 9= 31, = 1,=1.0 A The total current drawn by the circuit is Ignal = 1, + 1, + I, = 3.0 + 1.3 + 1.0 = 5.3 A total The total resistance is given by 1 R,' R, 1 1 + 1 1 +=+ 7'9 1 R = 1.70 R, 3 total totat Note that we could have obtained the total current drawn by the circuit using the total resistance and the Ohm's law in the following manner: V = R - 9 V= (1.7 0)(Ia) = Ie = 5.3 A total* total total total
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