# In step 1 of the above figure r eq 12 ω is returned

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In Step 1 of the above figure, R eq = 12 Ω is returned to R 1 = 6 Ω and R 234 = 6 Ω in series. Both resistors must have the same 2.0 A current as R eq . We then use Ohm’s law to find Δ V R1 = (2 A)(6 Ω ) = 12 V Δ V R234 = (2 A)(6 Ω ) = 12 V As a check, 12 V + 12 V = 24 V, which was Δ V of the R eq resistor. In Step 2, the resistance R 234 is returned to R 2 and R 34 in parallel. Both resistors must have the same Δ V = 12 V as the resistor R 234 . Then from Ohm’s law,

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R2 R34 12 V 12 V 0.80 A 1.2 A 15 10 I I = = = = Ω Ω As a check, I R2 + I R34 = 2.0 A, which was the current I of the R 234 resistor. In Step 3, R 34 is returned to R 3 and R 4 in series. Both resistors must have the same 1.2 A as the R 34 resistor. We then use Ohm’s law to find ( Δ V ) R3 = (1.2 A)(6 Ω ) = 7.2 V ( Δ V ) R4 = (1.2 A) (4 Ω ) = 4.8 V As a check, 7.2 V + 4.8 V = 12 V, which was Δ V of the resistor R 34 . Resistor Potential difference (V) Current (A) 6 Ω left 15 Ω 6 Ω right 4 Ω 12 12 7.2 4.8 2 0.80 1.2 1.2
32.61. Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.61 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 diagram, 12 V 2 A 6 6 I = = = Ω Ω E Thus, the current through the battery is 2 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 Δ V . In Step 1, the 6 Ω resistor is returned to a 3 Ω and 3 Ω resistor in series. Both resistors must have the same 2 A current as the 6 Ω resistance. We then use Ohm’s law to find Δ V 3 = (2 A)(3 Ω ) = 6 V As a check, 6 V + 6 V = 12 V, which was Δ V of the 6 Ω resistor. In Step 2, one of the two 3 Ω resistances is returned to the 4 Ω , 48 Ω , and 16 Ω resistors in parallel. The three resistors must have the same Δ V = 6 V. From Ohm’s law, 4 6 V 1.5 A 4 I = = Ω 48 6 V 1 A 48 8 I = = Ω 16 6 V 3 A 16 8 I = = Ω Resistor Potential difference (V) Current (A) 3 Ω 4 Ω 48 Ω 16 Ω 6 6 6 6 2 1.5 1.2 3.8

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32.62. Model: The battery and the connecting wires are ideal. Visualize: The figure shows how to simplify the circuit in Figure P32.62 using the laws of series and parallel resistances. We will reverse the procedure and “build up” the circuit using the loop law and junction law to find the current and potential difference of each resistor. Solve: Having found R eq = 12 Ω , the current from the battery is I = (24 V)/(12 Ω ) = 2.0 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 Δ V . In Step 1 of the above figure, the 12 Ω resistor is returned to 4 Ω and 8 Ω resistors in series. Both resistors must have the same 2.0 A as the 12 Ω resistor. We use Ohm’s law to find Δ V 4 = 8 V and Δ V 8 = 16 V. As a check, 8 V + 16 V = 24 V, which was Δ V of the 12 Ω resistor. In Step 2, the 8 Ω resistor is returned to the 12 Ω and 24 Ω resistors in parallel. Both resistors must have the same
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• Winter '10
• E.Salik
• Resistor, Potential difference, Ω, Series and parallel circuits

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