Parallel and series connection of conductors: theory, formulas, connection and calculation of current

Almost everyone who was engaged in electricians had to solve the issue of parallel and serial connection of circuit elements. Some solve the problems of parallel and series connection of conductors by the "poke" method, for many "fireproof" garland is an inexplicable, but familiar axiom. Nevertheless, all these and many other similar questions are easily solved by the method proposed at the very beginning of the 19th century by the German physicist Georg Ohm. The laws discovered by him are still in effect, and almost everyone can understand them.

Basic electrical quantities of the circuit

In order to find out how this or that connection of conductors will affect the characteristics of the circuit, it is necessary to determine the values ​​that characterize any electrical circuit. Here are the main ones:

• Electrical voltage, according to scientific definition, is a potential difference between two points of an electrical circuit. Measured in volts (V). Between the terminals of a household outlet, for example, it is equal to 220 V, on a battery the voltmeter will show 1.5 V, and the charger of your tablet or smartphone gives out 5 V. Voltage can be alternating and constant, but in our case it is insignificant.
• Electric current is the orderly movement of electrons in an electrical circuit. The closest analogy is the flow of water in a pipeline. Measured in amperes (A). If the circuit is not closed, the current cannot exist.
• Electrical resistance. The value is measured in ohms (ohms) and characterizes the ability of a conductor or electrical circuit to resist the passage of electric current. If we continue the analogy with the plumbing, then a new smooth pipe will have little resistance, clogged with rust and slag - high.
• Electric power. This value characterizes the rate of conversion of electrical energy into any other and is measured in watts (W). A 1000-watt boiler will boil water faster than a hundred-watt one, a powerful lamp shines brighter, etc.

Mutual dependence of electrical quantities

Now you need to decide, as all of the above values ​​depend on one another. The dependency rules are simple and boil down to two basic formulas:

• I = U / R.
• P = I * U.

Here I is the current in the circuit in amperes, U is the voltage supplied to the circuit in volts, R is the resistance of the circuit in ohms, P is the electrical power of the circuit in watts.

Suppose we are dealing with a simple electrical circuit consisting of a power supply with voltage U and a conductor with resistance R (load).

Since the circuit is closed, current I flows through it. How big will it be? Based on the above formula 1, to calculate it, we need to know the voltage developed by the power supply and the load resistance. If we take, for example, a soldering iron with a 100 Ohm coil resistance and connect it to a 220 V lighting socket, then the current through the soldering iron will be:

220/100 = 2.2 A.

What is the power of this soldering iron? Let's use formula 2:

2.2 * 220 = 484 W.

A good soldering iron turned out, powerful, most likely two-handed. In the same way, operating with these two formulas and transforming them, you can find out the current through power and voltage, voltage across current and resistance, etc. How much, for example, does a 60 W light bulb consume in your table lamp:

60/220 = 0.27 A or 270 mA.

Resistance of the lamp spiral in operation:

220 / 0.27 = 815 ohms.

Multiple conductor circuits

All the cases discussed above are simple - one source, one load. But in practice, there can be several loads, and they are also connected in different ways. There are three types of load connection:

1. Parallel.
2. Consistent.
3. Mixed.

Parallel connection of conductors

The chandelier has 3 lamps, each with 60 watts. How much does a chandelier consume? That's right, 180 watts. First, we quickly calculate the current through the chandelier:

180/220 = 0.818 A.

And then her resistance:

220 / 0.818 = 269 ohms.

Before that, we calculated the resistance of one lamp (815 Ohm) and the current through it (270 mA). The resistance of the chandelier turned out to be three times lower, and the current - three times higher. And now it's time to take a look at the three-arm lamp diagram.

Diagram of a chandelier with three lamps

All lamps in it are connected in parallel and connected to the network. It turns out that when three lamps are connected in parallel, the total load resistance has decreased threefold? In our case, yes, but it is private - all lamps have the same resistance and power. If each of the loads has its own resistance, then a simple division by the number of loads is not enough to calculate the total value. But even here there is a way out - just use this formula:

1 / Rtot. = 1 / R1 + 1 / R2 +… 1 / Rn.

For ease of use, the formula can be easily converted:

Rtot. = (R1 * R2 *… Rn) / (R1 + R2 +… Rn).

Here Rtot. - the total resistance of the circuit when the load is connected in parallel. R1… Rn - resistances of each load.

Why the current increased when you connected three lamps in parallel instead of one is easy to understand - after all, it depends on the voltage (it remained unchanged) divided by the resistance (it decreased). It is obvious that the power in parallel connection will increase in proportion to the increase in current.

Serial connection

Now it's time to figure out how the parameters of the circuit will change if the conductors (in our case, the lamps) are connected in series.

Series connected load

The calculation of resistance in series connection of conductors is extremely simple:

Rtot. = R1 + R2.

The same three sixty-watt lamps connected in series will already amount to 2445 ohms (see. calculations above). What will be the consequences of increasing the resistance of the circuit? According to formulas 1 and 2, it becomes quite clear that the power and current strength will drop when the conductors are connected in series. But why are all the lamps burning dim now? This is one of the most interesting properties of daisy chaining and is widely used. Let's take a look at a garland of three lamps that are familiar to us, but connected in series.

Series connection of three lamps in a garland

The total voltage applied to the entire circuit remained 220 V. But it was divided between each of the lamps in proportion to their resistance! Since we have lamps of the same power and resistance, the voltage is divided equally: U1 = U2 = U3 = U / 3. That is, three times less voltage is now applied to each of the lamps, which is why they glow so dimly. Take more lamps - their brightness will drop even more. How to calculate the voltage drop across each of the lamps if they all have different resistances? For this, the four formulas given above are sufficient. The calculation algorithm will be as follows:

1. Measure the resistance of each of the lamps.
2. Calculate the total resistance of the circuit.
3. From the total voltage and resistance, you calculate the current in the circuit.
4. Based on the total current and resistance of the lamps, you calculate the voltage drop across each of them.

Want to consolidate your knowledge? Solve a simple problem without looking at the answer at the end:

You have at your disposal 15 miniature bulbs of the same type, designed for a voltage of 13.5 V. Is it possible to make a Christmas tree garland out of them, connected to a regular outlet, and if so, how?

Mixed connection

With the parallel and serial connection of conductors, you, of course, easily figured out. But what if you have something like this in front of you?

Mixed connection of conductors

How to determine the total resistance of a circuit? To do this, you need to split the circuit into several sections. The above construction is quite simple and there will be two sections - R1 and R2, R3. First, you calculate the total resistance of the parallel connected elements R2, R3 and find Rtot. 23. Then calculate the total resistance of the entire circuit consisting of R1 and Rtot.23 connected in series:

• Rtot. 23 = (R2 * R3) / (R2 + R3).
• Rchain = R1 + Rtotal 23.

The problem is solved, everything is very simple. And now the question is a little more complicated.

Complex mixed connection of resistances

How to be here? Likewise, you just need to show some imagination. Resistors R2, R4, R5 are connected in series. We calculate their total resistance:

Rtot. 245 = R2 + R4 + R5.

Now, in parallel to Rtotal. 245, connect R3:

Rtot.2345 = (R3 * Rtot.245) / (R3 + Rtot.245).

Well, then everything is obvious, since there are R1, R6 and the Rtotal 2345 we found, connected in series:

Rchains = R1 + Rtotal 2345 + R6.

That's all!

The answer to the Christmas tree garland problem

The lamps have an operating voltage of only 13.5 V, and in a 220 V outlet, so they must be connected in series.

Since the lamps are of the same type, the mains voltage will be divided equally between them and on each bulb there will be 220/15 = 14.6 V. The lamps are designed for a voltage of 13.5 V, therefore, although such a garland will work, it will burn out very quickly. To implement the idea, you need a minimum of 220 / 13.5 = 17, or better 18-19 bulbs.

Scheme of a Christmas tree garland of miniature incandescent lamps