Ohmâ€™s Law and the Combination of Resistors
Ohm's Law If / is the current through a given resistor and A V is the potential across the resistor, then

Two resistors (R_{1} = 10 Î© and R_{2} = 20 Î©) are connected in series with a potential source (9 volts). What is the current through resistor 1?
First, letâ€™s DRAW A DIAGRAM of the circuit (Figure 157).
We label the lower wire 0 volts. There is 9volt jump across the voltage source, so the upper wire is labeled 9 volts. We label the wire between the two resistors with the potential V_{m}. The current is the same through both resistors and the source.
The equivalent water course is shown in Figure 158 in which the current in the top trough is the same as the current through both waterfalls and through the bottom trough.
Figure 157  Figure 158 
If several resistors are in series, then the same current flow through all of them.
9V â€“ V_{m} = IR_{1}
V_{m} â€“ 0 = IR_{2}
9V â€“ IR_{2} = IR_{1}
9V = I(R_{1} + R_{2})
I = 9V/(R_{1} + R_{2})
= 9V/(10 Î© + 20 Î©)
= 0.3A
In the circuit above, what is the potential drop across resistor 1?
If we apply Ohmâ€™s law to resistor 1, then we have
Î”V_{1} = I_{1}R_{1}
= (0.3A)(10 Î©)
= 3V
Generally, we do not have to go through as much trouble as we did in Example 1 because there are two rules for combining resistors:
If several resistors (/?,. /?,, and so on) arc in series, then we can replace them with one resistor whose resistance is the sum

If several resistors (R_{1}, R_{2â€¦} and so on) are in parallel then we can replace them with one resistor with resistance R_{t}. where

In the circuit shown (Figure 159), we have R_{1} = 20 Î©, R_{2} = 30 Î©, and R_{3} = 60 Î©, and the potential source is 12 volts.
 What is the current through resistor R2?
 What is the current through the potential source?
Figure 159
 First, letâ€™s label the negative terminal of the source 0 V and the positive terminal 12 V. The potential all along the left wire is 0 volts, and along the right wire is 12 volts (Figure 1510).
This gives us the potential drop across R_{2}:
 Note that the current splits into three parts through the resistors. To obtain the total current, we need to combine resistors to obtain an equivalent circuit (Figure 1511):
Figure 1510  Figure 1511 
Figure 1512 shows the analogous watercourse.
Figure 1512
 Label the lower end of the battery 0 V.
 Label other wires with voltages.
 Label the currents going through the wires.
 Combine resistors.
 Apply Ohmâ€™s law.
Consider the circuit shown in Figure 1513, where R_{1} = 1 Î©, R_{2} = R_{3} = 4 Î©, and V_{cell} = 6 V.
 What is the current going through resistor 1?
 How does I_{1} change if points A and B are connected with a wire?
Figure 1513
 First we draw in the current and voltages (Figure 1514), but this does not seem to get anywhere, because we do not know a potential difference across resistor 1 to apply Ohmâ€™s law. (Many students make the mistake here of using 6 volts and 1 Î© in Ohmâ€™s law, but the potential difference across resistor 1 is not 6 volts.)
Figure 1514
Figure 1515  Figure 1516 
 If we connect points A and B by a wire, then they have the same potential, and this is shown in Figure 1517. No current flows through resistors 2 and 3, and resistor 1 has the full 6 V across it. The current through it (by Ohmâ€™s law) is 6 A.
Figure 1517