The bulbs in the string of figure :

**Light bulbs in series, with an ammeter (A) in the circuit.**

being all the same, each get the same amount of voltage from the source. If there are a dozen bulbs in a 120-V circuit, each bulb has a potential difference of 10 V across it. This will remain true even if the bulbs are replaced with brighter or dimmer ones, as long as all the bulbs in the string are identical.

Look at the schematic diagram of following Figure

Each resistor carries the same current. Each resistance Rn has a potential difference En across it equal to the product of the current and the resistance of that particular resistor. The voltages En are in series, like cells in a battery, so they add together. What if the voltages across all the resistors added up to something more or less than the supply voltage, E? Then there would have to be a “phantom EMF” someplace, adding or taking away voltage.

But that’s impossible. Voltage cannot come out of nowhere!

Look at this another way. The voltmeter V in above figure shows the voltage E of the battery, because the meter is hooked up across the battery. The voltmeter V also shows the sum of the voltages En across the set of resistances, because it’s connected across the whole combination. The meter says the same thing whether you think of it as measuring the battery voltage E or as measuring the sum of the voltages En across the series combination of resistances. Therefore, E is equal to the sum of the voltages En.

How do you find the voltage across any particular resistance Rn in a circuit like the one in above figure Remember Ohm’s Law for finding voltage: E = IR. 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.

#### Problem 1

**In above figure, there are 10 resistors. Five of them have values of 10 Ω, and the other five have values of 20 Ω. The power source is 15-V dc. What is the voltage across any one of the 10-Ω resistors? Across any one of the 20-Ω resistors?**

First, find the total resistance: R = (10 × 5) + (20 × 5) = 50 + 100 = 150 Ω. Then find the current: I = E/R = 15/150 = 0.10 A. This is the current through each of the resistances in the circuit.

- If Rn = 10 Ω, then En = IRn = 0.1 × 10 = 1.0 V.

- If Rn = 20 Ω, then En = IRn = 0.1 × 20 = 2.0 V.

Let’s check to be sure all of these voltages add up to the supply voltage. There are five resistors with 1.0 V across each, for a total of 5.0 V; there are also five resistors with 2.0 V across each, for a total of 10 V. So the sum of the voltages across the resistors is 5.0 + 10 = 15 V.

#### Problem 2

**In the circuit of above figure, what will happen to the voltages across the resistances if one of the 20-Ω resistances is replaced with a short circuit?**

In this case the total resistance becomes R = (10 × 5) + (20 × 4) = 50 + 80 = 130 Ω. The current is therefore I = E/R = 15/130 = 0.12 A. This is the current at any point in the circuit, rounded off to two significant figures.

The voltage En across any of the 10-Ω resistances Rn is equal to IRn, which is 0.12 × 10 = 1.2 V.

The voltage En across any of the 20-Ω resistances Rn is equal to IRn, which is 0.12 × 20 = 2.4 V.

Checking the total voltage, add (5 × 1.2) + (4 × 2.4) = 6.0 + 9.6 = 15.6 V. This rounds off to 16 V when we cut it down to two significant figures.

#### Rounding-Off Bug

Compare the result for total voltage in Problem 2 with the result for total voltage in following figure :

What is going on here? Where does the extra volt come from in the second calculation? Certainly, shorting out one of the resistances cannot cause the battery voltage to change!

This is an example of what can happen when you round off to a certain number of significant figures after calculating the value of some parameter X in a circuit, then change a different parameter Y in the circuit, and finally calculate the value of X again, rounding off to the same number of significant digits as you did the first time. The discrepancy is the result of a “rounding off bug.”

If this bug bothers you (and it should), keep all the digits your calculator will hold while you go through the solution process for Problem 2. The current in the circuit, as obtained by means of a calculator that can show 10 digits, should come out as 0.115384615 A. When you find the voltages across all the resistances Rn, accurate to all these extra digits, and then add them up, you’ll get a final rounded-off voltage of 15 V.

This example shows why it is a good idea to wait until you get the final answer in a calculation, or set of calculations, involving a particular circuit before you round off to the allowed number of significant digits. Rounding-off bugs of the sort we have just seen can be more than mere annoyances. They are easy to overlook, but they can generate large errors in iterative processes involving calculations that are done over and over.