32 1 Solve From the circuit in Figure EX32 1 we see that 50 and 100 resistors are connected in series across the battery Another resistor of 75 is also connected across the battery 32 2 Solve that resistor is connected to resistor and a capacitor in parallel In Figure EX32 2 the positive terminal of the battery is connected to a resistor The other end of 32 3 Model Assume that the connecting wires are ideal Visualize Please refer to Figure EX32 3 Solve The current in the 2 resistor is 1 I 2 10 V 5 6 V 2 2 A I downward Using Kirchhoff s junction law we see that This current flows toward the junction that is downward I I1 I2 3 A 2 A 5 A 3 A to the left The current in the 5 resistor is 32 4 Model The batteries and the connecting wires are ideal Visualize Please refer to Figure EX32 4 Solve a Choose the current I to be in the clockwise direction If I ends up being a positive number then the current really does flow in this direction If I is negative the current really flows counterclockwise There are no junctions so I is the same for all elements in the circuit With the 9 V battery being labeled 1 and the 6 V battery being labeled 2 Kirchhoff s loop law is iV V bat 1 V R V bat 2 E 1 IR E 2 0 I E 1 E 2 R 9 V 30 6 V 0 10 A Note the signs Potential is gained in battery 1 but potential is lost both in the resistor and in battery 2 Because I is positive we can say that I 0 100 A flows from left to right through the resistor b The graph shows 9 V gained in battery 1 VR IR 3 V lost in the resistor and another 6 V lost in battery 2 The final potential is the same as the initial potential as required 32 5 Model Assume ideal connecting wires and an ideal battery for which Vbat Visualize Please refer to Figure EX32 5 We will choose a clockwise direction for I Note that the choice of the current s direction is arbitrary because with two batteries we may not be sure of the actual current direction The 3 V battery will be labeled 1 and the 6 V battery will be labeled 2 Solve a Kirchhoff s loop law going clockwise from the negative terminal of the 3 V battery is E V closed loop V i i V V R V bat 2 0 bat 1 3 V 18 I 6 V 0 I 0 5 A 9 V 18 Thus the current through the 18 resistor is 0 5 A Because I is positive the current is left to right i e clockwise b Assess The graph shows a 3 V gain in battery 1 a 9 V loss in the resistor and a gain of 6 V in battery 2 The final potential is the same as the initial potential as required 32 6 Visualize Please refer to Figure EX32 6 Define the current I as a clockwise flow Solve circuit element From Kirchhoff s loop law a There are no junctions so conservation of current tells us that the same current flows through each As we go around the circuit in the direction of the current potential is gained in the battery Vbat and potential is lost in the resistors Vresistor IR The loop law is batE 15 V Vi Vbat V10 V20 0 0 batE IR1 IR2 batE I R1 R2 I E bat R R 2 1 15 V 30 0 50 A Now that we know the current we can find the potential difference across each resistor V10 IR1 0 50 A 10 5 0 V V20 IR2 0 50 A 20 10 0 V b 32 7 Model The 1500 W rating is for operating at 120 V Solve The hair dryer dissipates 1500 W at VR 120 V Thus the hair dryer s resistance is The current in the hair dryer when it is used is given by Ohm s law R V R P R 2 2 120 V 1500 W 9 60 I V R R 120 V 9 60 12 5 A 32 8 Model Assume ideal connecting wires and an ideal battery Visualize Please refer to Figure EX32 8 Solve The power dissipated by each resistor can be calculated from Equation 32 14 PR I2R provided we can find the current through the resistors Let us choose a clockwise direction for the current and solve for the value of I by using Kirchhoff s loop law Going clockwise from the negative terminal of the battery V i i Vbat VR1 VR2 0 12 V IR1 IR2 0 I 12 V R R 2 1 12 12 V 18 2 5 A 0 40 A The power dissipated by resistors R1 and R2 is P R1 2 I R 1 0 40 A 12 2 1 92 W P R2 2 I R 2 0 40 A 18 2 2 9 W 32 9 Model The 100 W rating is for operating at 120 V Solve A standard bulb uses V 120 V We can use the power dissipation to find the resistance of the filament P R 2V R 2 V P 2 120 V 100 W 144 But the resistance is related to the filament s geometry R L L 2 r A r L R 7 9 0 10 144 m 0 070 m 1 18 10 m 11 8 m 5 The filament s diameter is d 2r 23 6 m 32 10 Solve Making use of the conversions 1 J 1 s 1 W and 1 kW 1000 J s we can write 1 kWh 1000 J s 3600 s 3 6 106 J 32 11 Solve a The average power consumed by a typical American family is P avg 1000 1000 kWh month kWh 30 24 h 1000 720 kW 1 389 kW Because P V I with V as the voltage of the power line to the house I avg P avg V 1 389 kW 1389 W 120 V 120 V 11 6 A R avg 2 V P avg 2 120 V 1389 W 10 4 b Because P V 2 R 32 12 Solve The cost of running the waterbed 35 of the time for a year is 0 35 450 W 365 days 24 hr day kW 0 11 1000 W kW hr 152 32 13 Visualize Please refer to Figure EX32 13 Solve The three resistors are in series The total resistance of this combination is Req R 50 R 2R 50 Thus Req 50 as long as R 0 32 14 Model Assume ideal connecting wires and an ideal battery Solve As shown in Figure EX32 14 a potential difference of 5 0 V causes a current of 100 mA through the three series resistors The situation is the same if we replace the …
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