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How to find the resistance of serial and parallel circuits

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Series Resistance

We take three constant resistances R1, R2 and R3 and connect them to the circuit so that the end of the first resistance R1 is connected to the beginning of the second resistance R 2, the end of the second - to the beginning of the third R 3, and let the conductors lead to the beginning of the first resistance from the current source (Fig. 1).

This connection of resistances is called serial. It is obvious that the current in such a circuit will be the same at all its points.


Fig 1. Series Resistance

How to determine the total resistance of a circuit if we already know all the resistances included in it in series? Using the position that the voltage U at the terminals of the current source is equal to the sum of the voltage drops in the sections of the circuit, we can write:

U1 = IR1 U2 = IR2 and U3 = IR3

IR = IR1 + IR2 + IR3

Putting brackets on the right-hand side of equality I, we obtain IR = I (R1 + R2 + R3).

Having now divided both sides of the equality by I, we finally have R = R1 + R2 + R3

Thus, we came to the conclusion that with a series connection of resistances, the total resistance of the entire circuit is equal to the sum of the resistances of individual sections.

We verify this conclusion with the following example. Take three constant resistances whose values ​​are known (for example, R1 = 10 Ohm, R 2 = 20 Ohm and R 3 = 50 Ohm). We connect them in series (Fig. 2) and connect them to a current source whose emf is 60 V (neglect the internal resistance of the current source).


Fig. 2. An example of a series connection of three resistances

We calculate what indications should be given by the devices turned on, as shown in the diagram, if you close the circuit. We determine the external resistance of the circuit: R = 10 + 20 + 50 = 80 Ohms.

Find the current in the circuit according to Ohm's law: 60/80 = 0, 75 A

Knowing the current in the circuit and the resistance of its sections, we determine the voltage drop for each section of the circuit U 1 = 0.75 x 10 = 7.5 V, U 2 = 0.75 x 20 = 15 V, U3 = 0.75 x 50 = 37.5 V.

Knowing the voltage drop in the sections, we determine the total voltage drop in the external circuit, i.e., the voltage at the terminals of the current source is U = 7.5 + 15 + 37.5 = 60 V.

We have thus obtained that U = 60 V, i.e., the nonexistent equality of the EMF of the current source and its voltage. This is explained by the fact that we neglected the internal resistance of the current source.

Having now closed the key switch K, we can verify from the instruments that our calculations are approximately correct.

Parallel connection of resistances

We take two constant resistances R1 and R2 and connect them so that the beginnings of these resistances are included in one common point a, and the ends in another common point b. After connecting the points a and b with the current source, we get a closed electric circuit. This connection of resistances is called a parallel connection.


Fig 3. Parallel connection of resistances

Let us trace the current flow in this circuit. From the positive pole of the current source through the connecting conductor, the current will reach point a. At point a, it branches, since here the chain itself branches out into two separate branches: the first branch with resistance R1 and the second with resistance R2. Denote the currents in these branches by I1 and I 2, respectively. Each of these currents will go along its branch to point b. At this point, the currents merge into one common current, which will come to the negative pole of the current source.

Thus, with a parallel connection of resistances, a branched chain is obtained. Let's see what will be the ratio between the currents in our circuit.

We turn on the ammeter between the positive pole of the current source (+) and point a and notice its readings. Then, having included the ammeter (shown as a dashed line in the figure) into the wire connecting point b to the negative pole of the current source (-), we note that the device will show the same current strength.

This means that the current strength in the circuit before its branching (to point a) is equal to the current strength after branching the circuit (after point b).

We will now include an ammeter alternately in each branch of the circuit, remembering the readings of the device. Let the ammeter show the current strength I1 in the first branch, and I 2 in the second branch. Adding these two ammeter readings, we get the total current equal in value to the current I before branching (to point a).

Consequently, the strength of the current flowing to the branch point is equal to the sum of the currents flowing from this point. I = I1 + I2 Expressing this with the formula, we get

This ratio, which is of great practical importance, is called the law of a branched chain.

Let us now consider what the relationship between the currents in the branches will be.

Turn on a voltmeter between points a and b and see what it shows us. Firstly, the voltmeter will show the voltage of the current source, since it is connected, as can be seen from Fig. 3, directly to the terminals of the current source. Secondly, the voltmeter will show the voltage drops U1 and U2 at the resistances R 1 and R2, since it is connected to the beginning and end of each resistance.

Therefore, with a parallel connection of the resistances, the voltage at the terminals of the current source is equal to the voltage drop at each resistance.

This gives us the right to write that U = U1 = U2,

where U is the voltage at the terminals of the current source, U 1 is the voltage drop at the resistance R 1, U2 is the voltage drop at the resistance R2. Recall that the voltage drop in a section of a circuit is numerically equal to the product of the current flowing through this section and the resistance of the section U = IR.

Therefore, for each branch you can write: U1 = I1R1 and U2 = I2R2, but since U 1 = U2, then I1R1 = I2R2.

Applying the rule of proportion to this expression, we obtain I1 / I2 = U2 / U1, i.e., the current in the first branch will be so many times (or less) the current in the second branch, how many times the resistance of the first branch is less (or more) the resistance of the second branches.

So, we have come to the important conclusion that in the case of parallel connection of the resistances, the total current of the circuit branches out into currents inversely proportional to the values ​​of the resistance of the parallel branches. In other words, the greater the resistance of the branch, the lower the current flowing through it, and, conversely, the lower the resistance of the branch, the greater the current flowing through this branch.

We will verify the correctness of this dependence in the following example. We assemble a circuit consisting of two parallel-connected resistances R1 and R 2 connected to a current source. Let R1 = 10 Ohms, R2 = 20 Ohms and U = 3 V.

We first calculate what the ammeter included in each branch will show us:

I1 = U / R1 = 3/10 = 0.3 A = 300 mA

I 2 = U / R 2 = 3/20 = 0.15 A = 150 mA

Total current in the circuit I = I1 + I2 = 300 + 150 = 450 mA

Our calculation confirms that when the resistors are connected in parallel, the current in the circuit branches inversely with the resistances.

Indeed, R1 == 10 Ohms is half as much as R 2 = 20 Ohms, while I1 = 300 mA is twice as much as I2 = 150 mA. The total current in the circuit I = 450 mA branched into two parts so that most of it (I1 = 300 mA) went through a smaller resistance (R1 = 10 Ohms), and a smaller part (R2 = 150 mA) through a larger resistance (R 2 = 20 ohms).

This branching of current in parallel branches is similar to the flow of fluid through pipes. Imagine a pipe A, which at some point branches into two pipes B and C of different diameters. Since the diameter of the pipe B is larger than the diameter of the pipes B, more water will pass through the pipe B at the same time than through the pipe B, which has a greater resistance to the flow of water.

Fig. four . Less water will pass through a thin pipe in the same period of time than through a thick

We now consider what the total resistance of the external circuit, consisting of two parallel-connected resistances, will be equal to.

By this common resistance of the external circuit, it is necessary to understand such a resistance, which could be replaced at a given voltage of the circuit, both parallel-connected resistances, without changing the current before branching. This resistance is called equivalent resistance.

Let's return to the circuit shown in fig. 3 and see what the equivalent resistance of two parallel-connected resistances will be equal to. Applying Ohm's law to this circuit, we can write: I = U / R, where I is the current in the external circuit (to the branch point), U is the voltage of the external circuit, R is the resistance of the external circuit, i.e., equivalent resistance.

Similarly, for each branch, I1 = U1 / R1, I2 = U2 / R2, where I1 and I 2 are the currents in the branches, U 1 and U2 are the voltage on the branches, R1 and R2 are the resistance of the branches.

According to the law of a branched chain: I = I1 + I2

Substituting the current values, we obtain U / R = U1 / R1 + U2 / R2

Since with parallel connection U = U1 = U2, we can write U / R = U / R1 + U / R2

Putting U on the right-hand side of the equality in brackets, we get U / R = U (1 / R1 + 1 / R2)

Having now divided both sides of the equality by U, we finally have 1 / R = 1 / R1 + 1 / R2

Remembering that the inverse of resistance is called the conductivity, we can say that in the resulting formula 1 / R is the conductivity of the external circuit, 1 / R1 is the conductivity of the first branch, 1 / R2 is the conductivity of the second branch.

Based on this formula, we conclude: with a parallel connection, the conductivity of the external circuit is equal to the sum of the conductivities of the individual branches.

Therefore, in order to determine the equivalent resistance of the resistances connected in parallel, it is necessary to determine the conductivity of the circuit and take the value opposite to it.

From the formula it also follows that the conductivity of the circuit is greater than the conductivity of each branch, which means that the equivalent resistance of the external circuit is less than the smallest of the resistances connected in parallel.

Considering the case of a parallel connection of resistances, we took the simplest circuit, consisting of two branches. However, in practice, there may be cases when a chain consists of three or more parallel branches. What to do in these cases?

It turns out that all the relations we have obtained remain valid for a circuit consisting of any number of parallel-connected resistances.

To verify this, consider the following example.

We take three resistances R1 = 10 Ohm, R2 = 20 Ohm and R3 = 60 Ohm and connect them in parallel. Determine the equivalent circuit resistance (Fig. 5).


Fig. 5. A chain with three parallel connected resistances

Applying the formula 1 / R = 1 / R1 + 1 / R2 for this chain, we can write 1 / R = 1 / R1 + 1 / R2 + 1 / R3 and, substituting the known quantities, we obtain 1 / R = 1/10 + 1 / 20 + 1/60

Let us add this fraction: 1 / R = 10/60 = 1/6, i.e., the conductivity of the circuit is 1 / R = 1/6 Therefore, the equivalent resistance is R = 6 Ohms.

Thus, the equivalent resistance is less than the smallest of the resistances included in parallel in the circuit, i.e., less than the resistance R1.

Now let's see if this resistance is really equivalent, that is, one that could replace 10, 20 and 60 Ohms connected in parallel, without changing the current strength until the branching of the circuit.

Suppose that the voltage of the external circuit, and therefore the voltage at the resistances R1, R2, R3, is 12 V. Then the current strength in the branches will be: I1 = U / R1 = 12/10 = 1, 2 A I 2 = U / R 2 = 12/20 = 1, 6 A I 3 = U / R1 = 12/60 = 0, 2 A

We obtain the total current in the circuit using the formula I = I1 + I2 + I3 = 1.2 + 0.6 + 0.2 = 2 A.

We check by the formula of Ohm's law whether a current with a force of 2 A is obtained in the circuit if instead of the three parallel-connected resistances known to us, one equivalent resistance of 6 Ohms is turned on.

I = U / R = 12/6 = 2 A

As you can see, the resistance we found R = 6 Ohms is really equivalent for this circuit.

This can also be seen on measuring instruments, if you assemble a circuit with the resistances we have taken, measure the current in the external circuit (before branching), then replace the parallel-connected resistances with one 6 Ohm resistance and measure the current again. The readings of the ammeter in both cases will be approximately the same.

In practice, parallel connections can also be found for which it is easier to calculate the equivalent resistance, i.e., without first determining the conductivities, immediately find the resistance.

For example, if two resistances R1 and R2 are connected in parallel, then the formula 1 / R = 1 / R1 + 1 / R2 can be converted as follows: 1 / R = (R2 + R1) / R1 R2 and, solving the equality with respect to R, get R = R1 x R2 / (R1 + R2), i.e., when two resistances are connected in parallel, the equivalent circuit resistance is equal to the product of the resistances connected in parallel divided by their sum.

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