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Kirchhoff s first law
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The physicist Gustav Robert Kirchhoff (1824-1887) was a researcher and experimentalist in electricity back in the time before radio, before electric lighting, and before much was understood about how currents flow.
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90 Direct-current circuit analysis Kirchhoff reasoned that current must work something like water in a network of pipes, and that the current going into any point has to be the same as the current going out. This is true for any point in a circuit, no matter how many branches lead into or out of the point. Two examples are shown in Fig. 5-5.
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In a network of water pipes that does not leak, and into which no water is added along the way, the total number of cubic feet going in has to be the same as the total volume going out. Water can t form from nothing, nor can it disappear, inside a closed system of pipes. Charge carriers, thought Kirchhoff, must act the same way in an electric circuit. This is Kirchhoff s First Law. An alternative name might be the law of conservation of current.
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Problem 5-11
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Refer to Fig. 5-5A. Suppose all three resistors have values of 100 , and that I1 2.0 A while I2 1.0 A. What is the battery voltage First, find the current I drawn from the battery. It must be 3.0 A; I I1 I2 2.0 1.0 3.0 A. Next, find the resistance of the whole combination. The two 100- resistors in series give a value of 200 , and this is in parallel with 100 . You can do the
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5-5 Kirchhoffs First Law. At A, the current into either X or Y is the same as the current out of that point: I = Il + I2. At B, the current into Z equals the current out of Z: Il + I2 = I3 + I4+ I5. Also see quiz questions 13 and 14.
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Kirchhoff s second law 91 calculations and find that the total resistance, R, across the battery, E, is 66.67 . Then E IR 66.67 3.0 200 volts. (Some battery.)
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Problem 5-12
In Fig. 5-5B, suppose each of the two resistors below point Z has a value of 100 , and all three resistors above Z are 10.0 . The current through each 100- resistor is 500 mA. What is the current through any of the 10.0- resistors, assuming it is equally distributed What is the voltage, then, across any of the 10. 0- resistors The total current into Z is 500 mA + 500 mA 1.00 A. This must be divided three ways equally among the 10- resistors. Therefore, the current through any one of them is 1.00/3 A 0.333 A 333 mA. The voltage across any one of the 10.0- resistors is found by Ohm s Law: E IR 0.333 10.0 3.33 V.
Kirchhoff s second law
The sum of all the voltages, as you go around a circuit from some fixed point and return there from the opposite direction, and taking polarity into account, is always zero. This might sound strange. Surely there is voltage in your electric hair dryer, or radio, or computer. Yes, there is, between different points. But no point can have an EMF with respect to itself. This is so simple that it s almost laughable. A point in a circuit is always shorted out to itself. What Kirchhoff really was saying, when he wrote his second law, is a more general version of the second and third points previously mentioned. He reasoned that voltage cannot appear out of nowhere, nor can it vanish. All the potential differences must balance out in any circuit, no matter how complicated and no matter how many branches there are. This is Kirchhoff s Second Law. An alternative name might be the law of conservation of voltage. Consider the rule you ve already learned about series circuits: The voltages across all the individual resistors add up to the supply voltage. Yes, they do, but the polarities of the EMFs across the resistors are opposite to that of the battery. This is shown in Fig. 5-6. It s a subtle thing. But it becomes clear when a series circuit is drawn with all the components, including the battery or other EMF source, in line with each other, as in Fig. 5-6.
Problem 5-13
Refer to the diagram of Fig. 5-6. Suppose the four resistors have values of 50, 60, 70 and 80 , and that the current through them is 500 mA. What is the supply voltage, E Find the voltages E1, E2, E3, and E4 across each of the resistors. This is done via Ohm s Law. In the case of E1, say with the 50- resistor, calculate E1 0. 500 50 25 V. In the same way, you can calculate E2 30 V, E3 35 V, and E4 40 V. The supply voltage is the sum El E2 E3 E4 25 30 35 40 V 130 V. Kirchhoff s Second Law tells us that the polarities of the voltages across the resistors are in the opposite direction from that of the supply in the above example.
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