Calculating current, voltage and resistance

Ohm's law indicates the relationship between current, voltage, and resistance, as we saw in Topic 7.

So far, we have only considered simple circuits consisting of just one resistor. How do we calculate the effect of multiple resistors?

Resistors in a loop add.

If two or more resistors are linked in a closed loop, the total resistance is just the sum of the individual resistances.

In this figure, the total resistance $R_{T} = R_{1} + R_{2} + R_{3}$

If $R_{1} = 100 Ω, R_{2} = 5000 Ω$ , and $R_{3} = 1 Ω$ , the total resistance is $100 + 5000 + 1 or 5101 Ω$ .

Circuit with three resistors in series

Resistors in parallel circuits form an equivalent resistance.

The total resistance for resistors in parallel is calculated using a formula. The total resistance is given as the equivalent resistance Req — i.e., the resistance of a single resistor that would have the same effect as all the resistors together. $R_{T}$ and Req may be used interchangeably.

The formula is:

$\frac{1}{R e q} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + ... . . + \frac{1}{R_{N}}$ where N is the number of resistors in parallel.

If the circuit just has two resistors in parallel, a little algebra shows that the general formula becomes:

$R e q = \frac{R_{1} x R_{2}}{R 1 + R_{2}}$

Similarly, for three resistors in parallel, the general formula becomes:

$R e q = \frac{R_{1} x R_{2} x R_{3}}{R_{1} + R_{2} + R_{3}}$

Circuit with two resistors in parallel

Say $R_{1} = 5 Ω, R_{2} = 10 Ω$ in the illustration at right.
Then the equivalent or total resistance is:

$R_{T} = \frac{(5 Ω) (10 Ω)}{5 Ω + 10 Ω} or \frac{50 Ω}{15 Ω} = 3.3 Ω$

A single resistor could be substituted for the two in parallel here, and have the same effect. Its value would be $R_{T}$ , or 3.3 $Ω$ . The circuit would then look like this:

Circuit with equivalent resistance to that of the circuit above.

If we were to use Ohm's law to calculate the current (as we did in Topic 7), we would use $R_{T}$ to find the total current.

Say $V = 9 V$ , $R_{1} = 5 Ω$ as above, $R_{2} = 10 Ω$ as above.
Then the total current $I = \frac{V}{R} = \frac{9 V}{3.3 Ω} = 2.7 A .$

Notice that the total or effective resistance in the circuit with parallel resistors is less than the value of any individual resistor. In this example,the total resistance is 3.3 $Ω$ , which is less than either the 5 $Ω$ or the 10 $Ω$ resistor.

It may seem counterintuitive: adding resistors in parallel DECREASES the total resistance.

As we saw above, adding resistors in series INCREASES the total resistance, as we would expect. Remember that resistors “resist” the flow of current. When resistors add in series, there is only one path for a given electron and it is “resisted” by all the resistors. If the same resistors add in parallel, there are multiple paths, each with less resistance than the sum of the three in series, so overall more electrons can get through.

A circuit can be understood as a combination of series and parallel circuits

First, find the equivalent resistance of the two resistors in parallel, $R_{2}$ and $R_{3}$ :
$R_{23} = \frac{R_{2} R_{3}}{R_{2} + R_{3}}$

Now that segment adds in series with the other two resistors $R_{1}$ and $R_{4}$ :

$R_{T} = R_{1} + R_{23} + R_{4}$

Let's use some numbers in this example.

Say $R_{1} = 1 Ω$ , $R_{2} = 5 Ω$ , $R_{3} = 8 Ω$ , and $R_{4} = 11 Ω$ .

First find $R_{23}$ :

$R_{23} = \frac{5 x 8}{5 + 8} = 3 A$

Now find the total resistance

$R_{T} = 1 + 3 + 11 = 15 Ω$

Circuit with two resistors in parallel, in series with two more resistors

Calculating currents

The current put out by a voltage source is neither created nor destroyed, so the sum of the current in each loop of a multi-loop circuit must add up to the total.

Remember this example?

As a check on our calculation of the total current of 2.7 A, we can figure out the currents in each parallel path. The currents in the two paths must add up to the total current.

In the path with $R_{1}$ , we have $I_{1} = \frac{V}{R} = \frac{9 V}{5 Ω} = 1.8 A$ .
In the path with $R_{2}$ , we have $I_{2} = \frac{V}{R} = \frac{9 V}{10 Ω} = 0.9 A$ .
Total current $I_{1} + I_{2} = 1.8 + 0.9 = 2.7 A$ . That is correct!

Calculating potential drops

The sum of the potential differences around a circuit must add up to zero. Remember the water-park analogy: the drop in water level from the top of the slide to the bottom must equal the rise in water level given by the pump.

The applied potential or emf in this circuit is LaTeX: V_1+V_2 - that's the total rise in potential. The potential drop across $R_{1}$ must be $(V_{1} + V_{2})$ ; the drop cancels the rise.
The potential drop across $R_{2}$ and R3 together must also be $(V_{1} + V_{2})$ .

If $V_{1} = 3 V$ and $V_{2} = 5 V$ ,
then the potential drop across $R_{1}$ is 8 V
and the potential drop across the leg of the circuit is also 8 V.

In summary: Ohm's law tells us everything we need to know.

Ohm's law can tell us the total current in the circuit if we know the applied voltage, the value of the resistors in the circuit, and how to add resistors. Knowing the current and resistance, we can use Ohm's law to figure out the voltage drop across a component. Knowing the potential difference and current between two points, we can use Ohm's law to find the resistance.

When we combine Ohm's law with the fact that the total current equals the sum of the current in each parallel loop, and the total potential drop equals the applied voltage, we can calculate the unknown quantities in a circuit.