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AP Physics Homework Solutions Unit 9: Electric Circuits Homework 5 Page 617 problems 5, 10, 12,14,16,18 18.5 (a) The equivalent resistance of the two parallel resistors is 1 1 1 Rp 4.12 7.00 10.0 Thus, Rab R 4 R p R9 4.00 4.12 9.00 17.1 (b) Iab 18.10 V ab 34.0 V R ab 17.1 1.99 A , so I4 I9 1.99 A Also, V p IabR p 1.99 A 4.12 8.18 V Then, I7 and I10 V p R7 V p R10 8.18 V 1.17 A 7.00 8.18 V 0.818 A 10.0 First, consider the parallel case. The resistance of resistor B is RB V B IB 6.0 V 3.0 2.0 A In the series combination, the potential difference across B is given by V B V battery V A 6.0 V-4.0 V 2.0 V The current through the series combination is then Is V B RB 2.0 V 2 A 3.0 3 and the resistance of resistor A is R A V A Is 4.0 V 6.0 23A 18.12 (a) The total current from a to b is equal to the maximum current allowed in the 100 series resistor adjacent to point a. This current has a value of Im ax Pm ax 25.0 W 0.500 A R 100 The total resistance is 1 1 1 R ab 100 100 50.0 150 100 100 Thus, V m ax Im axR ab 0.500 A 150 75.0 V (b) The power dissipated in the series resistor is P1 Im2 axR 25.0 W , and the power dissipated in each of the identical parallel resistors is P2 P3 Im ax 2 R 0.250 A 100 6.25 W 2 2 The total power delivered is P P1 P2 P3 25.0 6.25 6.25 W 37.5 W 18.14 The resistance of the parallel combination of the 3.00 and 1.00 resistors is 1 1 1 Rp 0.750 3.00 1.00 The equivalent resistance of the circuit connected to the battery is Req 2.00 R p 4.00 6.75 and the current supplied by the battery is I V 18.0 V 2.67 A R eq 6.75 The power dissipated in the 2.00- resistor is P2 I2 R2 2.67 A 2.00 14.2 W 2 and that dissipated in the 4.00- resistor is P4 I2 R 4 2.67 A 4.00 28.4 W 2 The potential difference across the parallel combination of the 3.00 and 1.00 resistors is V p IR p 2.67 A 0.750 2.00 V Thus, the power dissipation in these resistors is given by V p 2.00 V 2 2 P3 R3 3.00 V p 2.00 V 2 1.33 W 2 P1 and 18.16 R1 1.00 4.00 W Going counterclockwise around the upper loop, applying Kirchhoff’s loop rule, gives 15.0 V 7.00 I1 5.00 2.00 A 0 I1 or 15.0 V 10.0 V 0.714 A 7.00 From Kirchhoff’s junction rule, I1 I2 2.00 A 0 I2 2.00 A I1 2.00 A 0.714 A 1.29 A so Going around the lower loop in a clockwise direction gives E 2.00 I2 5.00 2.00 A 0 E 2.00 1.29 A 5.00 2.00 A 12.6 V or 18.18 Observe that the center branch of this circuit, that is the branch containing points a and b, is not a continuous conducting path, so no current can flow in this branch. The only current in the circuit flows counterclockwise around the perimeter of this circuit. Going counterclockwise around the this outer loop and applying Kirchhoff’s loop rule gives 8.0 V 2.0 I 3.0 I+12 V 10 I 5.0 I 0 or I 12 V 8.0 V 0.20 A 20 Now, we start at point b and go around the upper panel of the circuit to point a, keeping track of changes in potential as they occur. This gives Vab Va Vb 4.0 V+ 6.0 0 3.0 0.20 A 12 V 10 0.20 A 5.4 V Since V ab 0 , pointa is 5.4 V higherin potentialthan pointb