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CHAPTER 2 Electric Current. Ohm s Law
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Now let RT be the total resistance as seen by the battery in Fig. 28. Then, by Ohm s law, it has to be true that I V RT 31
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Since the left-hand sides of the last two equations are equal, the two right-hand sides are also equal. Setting the two right-hand sides equal, then canceling the Vs, gives 1 1 1 1 1 RT R1 R2 R3 Rn
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where RT is the total e ective resistance seen by the battery. In words, eq. (32) says that If n resistances are connected in parallel, in the manner of Fig. 28, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. The power input to each resistance in the parallel-connected circuit of Fig. 28 is (by eqs. (15), (16), (17)) found by any of the formulas
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2 Pn VIn V 2 =Rn In Rn
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where V is the battery voltage, and In is the current in resistor Rn . The total power output P of the battery is given by eqs. (25) and (26), where I is the battery current and where, now, RT is found by means of eq. (32). The following example will be helpful. Example
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In Fig. 29, the battery voltage is V 65 volts, and the values of the resistances, in ohms, are 38, 17, and 27, as shown.
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Fig. 29 In Fig. 29 we wish to nd the following values: (a) total resistance seen by the battery, (b) current measured by the ammeters shown in the gure, (c) power output of the battery, (d) power input to each resistor.
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CHAPTER 2 Electric Current. Ohm s Law
Solutions (a) Note that the three resistors are connected in parallel. Hence the resistance RT seen by the generator (battery) is, by eq. (32), 1 1 1 1 0:122 176 reciprocal ohms; RT 38 17 27 thus RT 1=0:122176 8:18489 ohms, answer. I battery current 65=RT 65=8:18489 7:94146 amperes, answer. I1 65=R1 65=38 1:710526 amperes, answer. I2 65=R2 65=17 3:823529 amperes, answer. I3 65=R3 65=27 2:407407 amperes, answer. (c) P VI 65 7:94146 516:195 watts, answer. (d) P1 VI1 65 1:710526 111:185 watts, answer. P2 VI2 65 3:823529 248:529 watts, answer. P3 VI3 65 2:407407 156:482 watts, answer. (b) You should verify using your calculator, that the answers to parts (c) and (d) satisfy eq. (26). Note: For the special case of two resistors in parallel, as in Fig. 30, eq. (32) gives the value 1=RT 1=R1 1=R2 which, after combining the two fractions together over the common denominator R1 R2 , then inverting both sides, becomes
RT
R1 R2 R1 R2
34
Fig. 30
Thus, the resistance seen looking into terminals a, b in Fig. 30 is given by eq. (34), being equal to the product of the two resistors, over their sum. The combination of two resistors in parallel occurs so often in practical work that eq. (34) should be memorized. Another special case sometimes encountered is that where the n resistors in Fig. 28 all have the same value of R ohms. In that case, eq. (32) becomes 1=RT n=R, which, after inverting both sides, becomes RT R n 35
Thus, in Fig. 28, if the n parallel resistances all have the same value of R ohms, then the battery sees a total resistance of R=n ohms. Problem 17 A battery having a constant terminal voltage of 18 volts is applied to ve parallelconnected resistance loads, as shown below. Resistance values in ohms. Find the following: (a) (b) resistance seen by battery, battery current,
CHAPTER 2 Electric Current. Ohm s Law
(c) (d) (e)
power output of battery, current in each resistor (check to see that eq. (28) is satis ed), power in each resistor (check to see that eq. (26) is satis ed).
Problem 18 A battery of constant voltage 15 volts is connected to two parallel-connected resistor loads of 25 ohms and 38 ohms. Find battery current and battery power. Problem 19 A battery of constant 12 volts is applied to sixteen 25-ohm resistors connected in parallel. Find (a) battery current, (b) current in each resistor, (c) battery power output, (d) power delivered to each resistor.
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