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Resistors in Series 

We've looked at Ohm's Law for a circuit containing a single resistor, and found it to be quite simple. But what happens if we have multiple resistors in the circuit? How can we deal with that, and still perform accurate calculations on the overall circuits?
On this page, we'll consider a circuit using two resistors, connected so that the same current flows through both, and the two resistors must share the applied voltage. This is known as a series connection.
In the circuit shown to the right, we see two resistors instead of one. To distinguish them without assigning fixed values to them, we designate them "R1" and "R2."
In this circuit, we don't know the specific voltage across each resistor, but we do know that the same current must be flowing through both of them. We will use that fact to help us with our calculations.
We have already said that the current must be the same everywhere in a series circuit. If it were not so, we would have to allow current to be created or destroyed at random, which is not possible. Therefore, the same current, I, flows through both resistors.
We can also apply Ohm's Law separately to each resistor, by noting that the voltage E1 across R1 must be:
E1 = I × R_{1}
Similarly,E2 = I × R_{2}
But we also know that the total battery voltage E = E_{1} + E_{2}. Therefore we can say that:
E = E_{1} + E_{2}
E = (I × R_{1}) + (I × R_{2})
E = I × (R_{1} + R_{2})
That last equation has a serious consequence throughout all of electronics: resistors in series add their values together to get the total resistance.
In addition, we can now solve for the total circuit current:
I = E ÷ (R_{1} + R_{2})
Additional resistors in series simply continue to add to the total circuit resistance.
In the circuit to the right, the total resistance R_{1} + R_{2} = 1.8k + 5.6k = 7.4k. Incidentally, the designation "k" specifies units of kilohms, or thousands of ohms. The "k" stands for kilo, meaning 1000. Most practical circuits involve resistances of this magnitude, so they are often specified this way.
In any case, we can now calculate the circuit current:
I = E ÷ (R_{1} + R_{2})
I = 9 ÷ (1.8k + 5.6k)
I = 9 ÷ 7.4k
I = 1.216 mA
We can now also calculate the voltage across each resistor:
E1 = I × R_{1}
E1 = 1.216 mA × 1.8k
E1 = 2.19 v.
and,E2 = I × R_{2}
E2 = 1.216 mA × 5.6k
E2 = 6.81 v.
Finally,E = E_{1} + E_{2}
E = 2.19 v. + 6.81 v.
E = 9 v.
Note that we can make our calculations directly and without converting units if we match milliAmperes (mA) with volts (v) and kilohms (k). This works because the changes in orders of magnitude cancel each other out. This is not true of all kinds of calculations, however, so you have to be careful of how you handle your units.
In addition, we did round off our results above. However, if you make the calculations with greater precision, you will find that the results still add up correctly.


 
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