B there are no junctions so the same current i flows

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(b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoff’s loop law in the counterclockwise direction and starting at the lower right corner, ∑Δ V i = 12 V I (12 Ω ) I (6 Ω ) 6 V – IR = 0 Note that the IR terms are all negative because we’re applying the loop law in the direction of current flow, and the potential decreases as current flows through a resistor. We can easily solve to find the unknown resistance R : 6 V I (18 Ω ) – IR = 0 ( ) ( )( ) 6 V 18 6 V 18 0.25 A 6 0.25 A I R I Ω Ω = = = Ω (c) The power is P = I 2 R = (0.25 A) 2 (6 Ω ) = 0.38 W. (d) The potential difference across a resistor is Δ V = IR , giving Δ V 6 = 1.5 V, and Δ V 12 = 3 V. Starting from the lower left corner, the graph goes around the circuit clockwise , opposite from the direction in which we applied the loop law. In this direction, we speak of potential as lost in the batteries and gained in the resistors.
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32.44. Model: Assume that the connecting wires are ideal but the battery is not ideal. Visualize: Solve: The figure shows a variable resistor R connected across the terminals of a battery that has an emf E and an internal resistance r . Using Kirchhoff’s loop law and starting from the lower left corner, + E Ir IR = 0 E = I ( r + R ) From the point in Figure P32.44 that corresponds to R = 0 Ω , E = (6 A)( r + 0 Ω ) = (6 A) r From the point that corresponds to R = 10 Ω , E = (3 A)( r + 10 Ω ) Combining the two equations, (6 A) r = (3 A)( r + 10 Ω ) 2 r = r + 10 Ω r = 10 Ω Also, E = (3 A)(10 Ω + 10 Ω ) = 60 V. Assess: With E = 60 V and r = 10 Ω , the equation E = I ( r + R ) is satisfied by all values of R and I on the graph in Figure P32.44.
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32.45. Model: The connecting wires are ideal, but the battery is not. Visualize: Please refer to Fig. P32.45. We will designate the current in the 5 Ω resistor I 5 and the voltage drop Δ V 5 . Similar designations will be used for the other resistors. Solve: Since the 10 Ω resistor is dissipating 40 W, 2 10 10 10 40 W P I R = = 10 10 10 40 W 2.0 A 10 P I R = = = Ω Δ V 10 = I 10 R 10 = (2.0 A)(10 Ω ) = 20 V The 20 Ω resistor is in parallel with the 10 Ω resistor, so they have the same potential difference: Δ V 20 = Δ V 10 = 20 V. From Ohm’s law, 20 20 20 20 V 1.0 A 20 V I R Δ = = = Ω The combined current through the 10 Ω and 20 Ω resistors first passes through the 5 Ω resistor. Applying Kirchhoff’s junction law at the junction between the three resistors, I 5 = I 10 + I 20 = 1.0 A + 2.0 A = 3.0 A ⇒ Δ V 5 = I 5 R 5 = (3.0 A)(5 Ω ) = 15 V Knowing the currents and potential differences, we can now find the power dissipated: P 5 = I 5 Δ V 5 = (3.0 A)(15 V) = 45 W P 20 = I 20 Δ V 20 = (1.0 A)(20 V) = 20 W
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32.46. Model: Assume that the connecting wires are ideal, but the battery is not. The battery has internal resistance. Also assume that the ammeter does not have any resistance. Visualize: Please refer to Figure P32.46. Solve: When the switch is open, E – Ir – I (5.0 Ω ) = 0 V E = (1.636 A)( r + 5.0 Ω ) where we applied Kirchhoff’s loop law, starting from the lower left corner. When the switch is closed, the current I comes out of the battery and splits at the junction. The current I = 1.565 A flows through the 5.0 Ω resistor and the rest ( I I ) flows through the 10.0 Ω resistor. Because the potential differences across the two
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