Current Through Series Resistances
Have you ever used those tiny holiday lights that come in strings? If one bulb burns out, the whole set of bulbs goes dark. Then you have to find out which bulb is bad and replace it to get the lights working again. Each bulb works with something like 10 V, and there are about a dozen bulbs in the string. You plug in the whole bunch, and the 120V utility mains drive just the right amount of current through each bulb.
In a series circuit such as a string of light bulbs, the current at any given point is the same as the current at any other point. An ammeter can be connected in series at any point in the circuit, and it will always show the same reading. This is true in any series dc circuit, no matter what the components actually are and regardless of whether or not they all have the same resistance.
If the bulbs in a string are of different resistances, some of them will consume more power than others. In case one of the bulbs burns out and its socket is shorted out instead of filled with a replacement bulb, the current through the whole chain will increase because the overall resistance of the string will go down. This will force each of the remaining bulbs to carry too much current. Another bulb will burn out before long as a result of this excess current. If it, too, is replaced by a short circuit, the current will be increased still further. A third bulb will blow out almost right away. At this point it would be wise to buy some new bulbs !
Voltages Across Series Resistances
In a series circuit, the voltage is divided up among the components. The sum total of the potential differences across each resistance is equal to the dc powersupply or battery voltage. This is always true, no matter how large or how small the resistances and whether or not they’re all the same value.
If you think about this for a moment, it’s easy to see why this is true. Look at the schematic diagram of Fig. 1211. Each resistor carries the same current. Each resistor R _{n} has a potential difference E _{n} across it equal to the product of the current and the resistance of that particular resistor. These E _{n} values are in series, like cells in a battery, so they add together. What if the E _{n} values across all the resistors added up to something more or less than the supply voltage E? Then there would be a “phantom emf” someplace, adding or taking away voltage. However, there can be no such thing. An emf cannot come out of nowhere.
Fig. 1211 . Analysis of voltage in a series dc circuit. See text for discussion.
Look at this another way. The voltmeter V in Fig. 1211 shows the voltage E of the battery because the meter is hooked up across the battery. The meter V also shows the sum of the E _{n} values across the set of resistors simply because the meter is connected across the set of resistors. The meter says the same thing whether you think of it as measuring the battery voltage E or as measuring the sum of the E _{n} values across the series combination of resistors. Therefore, E is equal to the sum of the E _{n} values.
This is a fundamental rule in series dc circuits. It also holds for common utility ac circuits almost all the time.
How do you find the voltage across any particular resistor R _{n} in a circuit like the one in Fig. 1211? Remember Ohm’s law for finding voltage: E = IR . The voltage is equal to the product of the current and the resistance. Remember, too, that you must use volts, ohms, and amperes when making calculations. In order to find the current in the circuit I , you need to know the total resistance and the supply voltage. Then I = E/R . First find the current in the whole circuit; then find the voltage across any particular resistor.

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