PHYS
Chapter 32

# Both resistors must have the same δ v 16 v as the 8

• Notes
• 83

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resistors in parallel. Both resistors must have the same Δ V = 16 V as the 8 Ω resistor. From Ohm’s law, ( ) ( ) 4 12 3 16 V 12 A I = Ω = and 2 24 3 A I = . As a check, I 12 + I 24 = 2.0 A, which was the current I of the 8 Ω

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resistor. In Step 3, the 12 Ω resistor is returned to the two 6 Ω resistors in series. Both resistors must have the same 4 3 A as the 12 Ω resistor. We use Ohm’s law to find Δ V 6 = 8 V and Δ V 6 = 8 V. As a check, 8 V + 8 V = 16 V, which was Δ V of the 12 Ω resistor. Finally, in Step 4, the 6 Ω resistor is returned to the 8 Ω and 24 Ω resistors in parallel. Both resistors must have the same Δ V = 8 V as the 6 Ω resistor. From Ohm’s law, I 8 = (8 V)/(8 Ω ) = 1 A and 1 24 3 A I = . As a check, 4 8 24 3 A I I + = , which was the current I of the 6 Ω resistor. Resistor Potential difference (V) Current (A) 4 Ω 8 2 6 Ω 8 1.3 8 Ω 8 1 Bottom 24 Ω 8 0.33 Right 24 Ω 16 0.66
32.63. Model: The batteries and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.63 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: From the last circuit in the figure and from Kirchhoff’s loop law, 12 V 3 V 1.0 A 9 I = = Ω Thus, the current through the batteries is 1.0 A. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. In Step 1 of the above figure, the 9 Ω resistor is returned to the 6 Ω and 3 Ω resistors in series. Both resistors must have the same 1.0 A current as the 6 Ω resistor. We use Ohm’s law to find Δ V 3 = (1.0 A)(3 Ω ) = 3.0 V Δ V 6 = 6.0 V As a check, 3.0 V + 6.0 V = 9 V, which was Δ V = (12 V – 3 V) = 9 V of the 9 Ω resistor. In Step 2, the 6 Ω resistor is returned to the 24 Ω and 8 Ω resistors in parallel. The two resistors must have the same potential difference Δ V = 6.0 V. From Ohm’s law, 7 6.0 V 3 A 8 4 I = = Ω 24 6.0 V 1 A 24 4 I = = Ω

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As a check, 0.75 A + 0.25 A = 1.0 A which was the current I of the 6 Ω resistor. In Step 3, the 8 Ω resistor is returned to the 3 Ω and 5 Ω (right) resistors in series, so the two resistors must have the same current of 0.828 A. We use Ohm’s law to find Δ V 3 = (3/4 A)(3 Ω ) = 9/4 V Δ V 4 = (3/4 A)(5 Ω ) = 15/4 V As a check, 9/4 V + 15/4 V = 24/4 V = 6.0 V, which was Δ V of the 8 Ω resistor. In Step 4, the 3 Ω resistor is returned to 4 Ω (left) and 12 Ω resistors in parallel, so the two must have the same potential difference Δ V = 9/4 V. From Ohm’s law, 4 9/4 V 9/16 A = 0.56 A 4 I = = Ω 12 9/4 V 9 A 0.19 A 12 48 I = = = Ω As a check, 0.56 A + 0.19 A = 0.75 A, which was the same as the current through the 3 Ω resistor. Resistor Potential difference (V) Current (A) 24 Ω 6.0 0.25 3 Ω 3.0 1.0 5 Ω 3.75 0.75 4 Ω 2.25 0.56 12 Ω 2.25 0.19
32.64.

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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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