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CHAPTER 2 Electric Current. Ohm s Law
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Secondly, if the battery is delivering a current I, the presence of Rb causes an internal voltage drop in the battery, equal to IRb , which subtracts from the useful voltage output of the battery, as explained in connection with Fig. 26, where x and y are the external terminals of the battery.
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Notice that the voltage drop across the internal resistance opposes or bucks the internal voltage generated by the battery, thereby reducing the useful voltage available at the battery terminals. This e ect is proportional to the current I, so that the more current we attempt to make the battery produce, the lower is the output voltage available at the battery terminals. This is an undesirable e ect because, ideally, we would like the battery voltage to remain constant, and not be a ected by changes in current. It follows, therefore, that a battery or other type of generator that must produce large values of current must be constructed to have very low internal resistance. If this is not done, the output voltage of the battery will vary widely with changes in output current, and the battery will be said to have poor voltage regulation. In our work here, we ll assume the batteries to have negligibly small values of internal resistance. In cases where this cannot be assumed, the internal resistance of the battery will be added in series with the battery, and its e ect included in calculations of current. The last point we wish to make is that circuits in which the direction of current ow does not change are called DIRECT-CURRENT or dc circuits. Since the direction of current ow through a battery does not change (unless it is being recharged ), a battery is an example of a dc generator, and thus the battery circuits in this and the next few sections are examples of dc circuits. Later in our work we ll take up the important case where the polarity of the generator periodically reverses, providing what is called alternating or ac circuits. Problem 15 In Fig. 27, a battery of constant 48 volts is applied to ve series-connected resistors, as shown, the resistance values being in ohms. (a) (b) (c) What is the reading of meter M1 What is the reading of meter M2 Power output of the battery is watts
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CHAPTER 2 Electric Current. Ohm s Law
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Problem 16 In Fig. 27, calculate the power input to each resistor, then verify that eq. (26) gives the same answer as found in part (c) above.
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The Parallel Circuit
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A PARALLEL circuit is one in which the battery current divides into a number of parallel paths. This is shown in Fig. 28, in which a battery, of constant emf V volts, delivers a current of I amperes to a load consisting of any number of n resistances connected in parallel.
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Fig. 28
The currents in the individual resistances are called the branch currents, and the battery current I is often called the line current. From inspection of Fig. 28 we see that, in a parallel circuit, the battery current I is equal to the sum of the branch currents, that is, in Fig. 28, I I1 I2 I3 In 28
Next, from Fig. 28 we see that the battery voltage V is applied equally to all n resistances; that is, the same voltage V is applied to all the parallel branches. Hence, by Ohm s law (eq. (11)), the individual branch currents in Fig. 28 have the values I1 V=R1 ; I2 V=R2 ; . . . ; In V=Rn 29
Upon substituting these values into the right-hand side of eq. (28) we have   1 1 1 1 I V R1 R2 R3 Rn
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