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Total power at parallel connection. Calculation of the resulting resistance in a series-parallel connection

This is a fairly simple topic, but very important. There are only two rules: with a series connection, the resistors are added, and in parallel, their conductivities are added, which, by definition, from chapter /, are the values inverse to the resistance (see Fig. 5.3). One can understand why the rules are exactly the same, if we consider the currents in both cases - when they are connected in series, the current I through the resistors is the same, therefore the voltage drops across them are added, which is equivalent to the summation of resistances. With parallel connection, on the contrary, the voltage drops are equal to U, and it is necessary to add currents, which is equivalent to the addition of conductivities. If you do not understand the above, then sit on rice. 5.3 with a pencil and paper and output the expressions of Ohm’s law for each of the cases - and everything will fall into place.

Fig. 5.3. Series and parallel connection of resistors

From these definitions, several practical rules emerge that are useful to learn:

With a serial connection:

A pair of resistors has a resistance that is always greater than the resistance of a resistor with a larger nominal value (the rule is “more large”);

If the ratings of the resistors are equal, then the total resistance is exactly twice as large as each rating;

If the values of the resistors differ many times, then the total resistance is approximately equal to the larger rating (a typical case was mentioned in Chapter 1: in the example in Fig. 1.4, we ignore the resistance of the wires, since it is much less than the resistance of the resistors);

With parallel connection:

A pair of resistors has a resistance that is always less than the resistance of a resistor with a lower nominal value (the rule “less than less”);

If the ratings of the resistors are equal, then the total resistance is exactly half the size of each rating;

If the values of the resistors differ many times, then the total resistance is approximately equal to the smaller rating (this can also be illustrated by the example of Fig. 1.4, where we ignore the presence of a voltmeter connected in parallel with R2, since its resistance is much greater than the resistance of the resistor).

Knowledge of these rules will help you quickly evaluate the scheme without doing algebra exercises and without using a calculator. Even if the ratio of resistances does not fall under these cases, the result can still be estimated “by eye” with sufficient accuracy. With a parallel connection, which is more difficult to calculate, for such an assessment it is necessary to estimate what fraction of the lower resistance is their arithmetic sum, their total resistance will decrease by about that time relative to the smaller one. This is easy to check: let one resistance be 3.3 kΩ, the second 6.8 kΩ. In accordance with the above, we would expect that the total resistance should be 30% less than 3.3 kΩ, that is, 2.2 kΩ (3.3 is about one third of the sum 3.3 + 6.8, that is total resistance should be less than 3.3, a third of this value, equal to 1.1 - as a result, we get 2.2). If we check the result obtained by such an estimate in the mind, by an exact calculation, then we will end up with a very close value of 2.22 kΩ.

In most cases, we don’t need such accuracy - remember that the resistances themselves are scattered at the nominal, and in most conventional schemes the tolerances on the nominal values of standard components can be quite large (at least in correctly composed circuits). If the scheme in some cases must still have some strictly defined parameters, then with the help of standard components you will not achieve it anyway - the parameters will “walk” (within tolerances, of course) from the wind blowing out of the vents, and in such In cases, precision resistors and capacitors should be used, and quartz resonators should be used during the timers circuits. But drawing a circuit so that it loses its working capacity from replacing a 1 kΩ resistor with a 1.1 kΩ resistor is not our method!

Calculation of resistance and power with parallel and series connection of resistors. (10+)

Calculation of parallel / serially connected resistors, capacitors and chokes

Parallel or serial connection (inclusion) is usually used in several cases. First of allif there is no resistor. Secondlyif there is a need to get a higher power resistor. ThirdlyIf you need to accurately select the nominal details, and install trimmer impractical for reasons of reliability. Most radio components have tolerances. To compensate for them, for example for a resistor, in series with a large resistor set smaller at times. Selection of this smaller resistor allows you to get exactly the desired resistance value.

Your attention a selection of materials:

Resistors

We connect in series

= +

= [The resistance of the first resistor, kΩ] * [Current, mA] ^ 2 / 1000

= [Resistance of the second resistor, kΩ] * [Current, mA] ^ 2 / 1000

It turns out that of the two 500 ohm resistors at 2 watts you can make one per 1 kΩ, 4 watts.

Turn on in parallel

1 / (1 / [The resistance of the first resistor, kΩ] + 1 / [Resistance of the second resistor, kΩ])

This formula is intuitive, and it can formally be derived from the following consideration. For a given voltage across the resistors, each of them independently conducts a current equal to the voltage divided by the resistance. The total resistance is equal to the voltage divided by the total current. In the formulas, the value of the voltage is happily reduced, and the resulting formula is obtained.

[Power dissipated by the first resistor, W] = [Voltage on resistors, V] ^ 2 / [The resistance of the first resistor, kΩ] / 1000

[Power dissipated by the second resistor, W] = [Voltage on resistors, V] ^ 2 / [The resistance of the first resistor, kΩ] / 1000

It turns out that of the two 500 ohm resistors at 2 watts you can make one at 250 ohms, 4 watts.

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Concepts and formulas

A series-parallel, or mixed, connection is a complex connection of three or more resistances. The resulting resistance for a mixed connection is calculated in stages using the formulas for calculating resistances for series and parallel connections.

Examples

First, we replace the parallel-connected resistances r2 and r3 with the resultant resistance r (2-3):

r (2-3) = (r2 ∙ r3) / (r2 + r3) = (10 20) / 30 = 6.6 ohms.

The resulting resistance of the entire circuit is r = r1 + r (2-3) = 5 + 6.6 = 11.6 ohms.

Fig. one.

2. What current flows through the circuit (Fig. 2) in the cases of open and closed P? How does the voltage on the resistance r2 in both cases?

Fig. 2

a) The switch is open. Resultant resistance of series-connected resistances r1 and r2

r (1-2) = r1 + r2 = 25 ohms.

Current I (1-2) = U / r (1-2) = 100/25 = 4 A.

Voltage drop across resistance r2

U2 = I (1-2) ∙ r2 = 4 ∙ 5 = 20 V.

b) The switch is closed. Resultant resistance of parallel-connected resistances r1 and r3

r (1-3) = (r1 ∙ r3) / (r1 + r3) = (20 ∙ 10) / (20 + 10) = 200/30 = 6.6 Ω.

The total resistance of the whole circuit is r = r (1-3) + r2 = 6.6 + 5 = 11.6 Ω.

Current I = U / r = 100 / 11.6 = 8.62 A.

The voltage drop across the resistance r2 in this case is: U2 = I ∙ r2 = 8.62 ∙ 5 = 43.25 V.

In the second case, the current increased as a result of connecting a parallel resistance R3. More current creates more on resistance r2.

3. What should be the rd so that two lamps connected in parallel for a voltage of 120 V and a current of 0.2 A can be included in the network with a voltage of U = 220 V (Fig. 3)?

Fig. 3

The voltage on the lamps must be equal to 120 V. The remaining voltage (100 V) falls on the additional resistance rd. Through the resistance rd passes the current of two lamps I = 0.4 A.

By Ohm's law, rd = Ud / I = 100 / 0.4 = 250 Ohm.

4. Electron tubes with a filament voltage of 1.2 V and a filament current of 0.025 and 0.05 A are connected in series to a 4.5 V DC source. What should be the additional resistance rd and to a lamp having a lower filament current (Fig. 4 )?

Fig. four.

The resistances in the circuit must be chosen so that the current of the second lamp I = 0.05 A flows. The voltage on the filaments of the electronic lamps will be 1.2 + 1.2 = 2.4 V. By subtracting this value from the battery voltage, we get the magnitude of the voltage drop across the additional resistance rd: Ud = 4.5-2.4 = 2.1 V.

Hence the additional resistance rd = (Ud) / I = 2.1 / 0.05 = 42 Ω.

Current 0.05 A does not have to flow all the way through the filament of the first electron tube. Half of this current (0.05-0.025 = 0.025 A) must pass through the shunt r. The voltage on the shunt is the same as that on the filament of the lamp, i.e. 1.2 V. Hence, the resistance of the shunt is: r = 1.2 / 0.025 = 48 ohm.

5. What are the resulting resistance of the circuit and the current in it in the circuit in fig. five?

Fig. five.

First of all, we define the resultant resistance of parallel-connected resistances:

r (1-2) = (r1 ∙ r2) / (r1 + r2) = (2 4) / (2 + 4) = 8/6 = 1.3 Ohm;

r (4-5) = (r4 ∙ r5) / (r4 + r5) = (15 ∙ 5) / (15 + 5) = 75/20 = 3.75 Ohms.

The resulting resistance of the circuit is:

r = r (1-2) + r3 + r (4-5) = 1.3 + 10 + 3.75 = 15.05 ohms.

The resulting current at a voltage of U = 90.5 V

I = U / r = 90.5 / 15.05 = 6 A.

Fig. 6

The resulting conductivity of parallel resistors included

1 / r (3-4-5) = 1 / r3 + 1 / r4 + 1 / r5 = 1/5 + 1/10 + 1/20 = 7/20 sim;

r (3-4-5) = 20/7 = 2.85 ohms.

The resistance of the circuit of r1 and r2 is equal to:

r (1-2) = r1 + r2 = 15 + 5 = 20 ohms.

The resulting conductivity and resistance between points A and B are respectively equal: 1 / rAB = 1 / r (3-4-5) + 1 / r (1-2) = 7/20 + 1/20 = 8/20 sim; rAB = 20/8 = 2.5 ohms.

The resulting resistance of the entire circuit is r = rAB + r6 = 2.5 + 7.5 = 10 ohms.

The resulting current I = U / r = 24/10 = 2.4 A.

The voltage between points A and B is equal to the source voltage U minus the voltage drop across the resistance r6

UAB = U-I ∙ r6 = 24- (2.4 ∙ 7.5) = 6 V.

This voltage is connected to the resistance r4, so the current passing through it will be equal to:

I4 = UAB / r4 = 6/10 = 0.6 A.

The resistances r1 and r2 have a common voltage drop UAB, therefore the current passing through the resistance r1 is equal to:

I1 = UAB / r (1-2) = 6/20 = 0.3 A.

Voltage drop across resistance r1

Ur1 = I1 ∙ r1 = 0.3 ∙ 15 = 4.5 V.

7. What are the resulting resistance and current in the circuit in fig. 7, if the source voltage U = 220 V?

Fig. 7

We start with the contour located to the right of nodes 3 and 3. The resistances r7, r8, r9 are connected in series, therefore

r (7-8-9) = r7 + r8 + r9 = 30 + 40 + 20 = 90 ohms.

Parallel to this resistance, resistance r6 is included, therefore the resultant resistance at node 3 and 3 (section a)

ra = (r6 ∙ r (7-8-9)) / (r6 + r (7-8-9)) = (20 ∙ 90) / (20 + 90) = 1800/110 = 16.36 Ω.

Consistently with resistance ra, resistances r4 and r5 are included:

r (4-5-a) = 10 + 20 + 16.36 = 46.3 ohms.

The resulting resistance at nodes 2 and 2 (section b)

rb = (r (4-5-a) ∙ r3) / (r (4-5-a) + r3) = (46.36 ∙ 30) / (466 + 30) = 1390.8 / 76, 36 = 18.28 ohms.

The resulting resistance of the entire circuit is r = r1 + rb + r2 = 40 + 18.28 + 10 = 68.28 ohms.

The resulting current I = U / r = 220 / 68.28 = 3.8 A.