Series and parallel connection of conductors summary presentation. Series and parallel connection of conductors

The experiment shows the relationship between current, voltage and resistance in a series connection. There are electrical circuits for serial connections and solutions to problems using these circuits.

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"Presentation for the lesson "Series connection of conductors."

Symbol

Name

Galvanic

Resistor

Ammeter

Voltmeter

Physical quantities and their letter designations.

Current strength

Voltage

Resistance

Ampere

Volt

Current strength

Voltage

Resistance

Physical quantities and instruments for their measurement.

Ammeter

Voltmeter

Current strength

Voltage

Georg Simon Ohm

Famous German physicist

Current (A)

I 1

I 2

Voltage (V)

U 1

U 2

Resistance (Ohm)

R 1

R 2

• With a series connection, the current strength in any part of the circuit is the same, i.e.

I = I 1 = I 2 .

The total voltage in the circuit in a series connection, or the voltage at the poles of the current source, is equal to the sum of the voltages in individual sections of the circuit:

U = U 1 +U 2

• The total resistance of the circuit when connected in series is equal to the sum of the resistances of the individual conductors:

R = R 1 + R 2 .

15 Ohm

20 Ohm

1. According to the diagram shown in Fig. 17, determine

ammeter readings and total resistance

in an electrical circuit, if R 1 = 5 Ohm, R 2 = 3 Ohm.

2. What are the ammeter readings and general

electrical circuit resistance,

shown in Fig. 18, If R 1 = 10 Ohm, R 2 = 2 Ohm?

3. According to the diagram shown in Fig. 21,

determine the ammeter readings and

resistance R2, if R1 = 4 Ohms.

Repetition

Chizhova Vera Alexandrovna

Physics and computer science teacher

MBOU Secondary School, Krasnoe village,

Nenets Autonomous Okrug.

• Speed ​​of charge movement along a conductor
• Charge passing through a conductor in 1s
• Denoted by ()
• Unit (A) ampere
• Measured by ammeter
• Depends on voltage and resistance (Ohm's Law)

• Voltage is the work done by an electric field to move a unit charge (1 C) along a conductor
• Identified by the letter (U)
• Measured by voltmeter
• Unit of measurement (V) volt

• The property of a conductor to interfere with the movement of charged particles along the conductor under the influence of an electric field
• Denoted by R
• Unit of measurement (Ohm)
• Depends on the physical properties of the conductor

Laws of series connection of conductors

• The current strength is the same in all parts of the circuit
• The total resistance of the circuit is equal to the sum of the resistances of the individual sections of the circuit
• The total stress is equal to the sum of the stresses in individual sections

• 1) You need to make a Christmas tree garland from light bulbs. designed for a voltage of 6 V, so that it can be connected to a 120 V network. How many light bulbs do you need to take for this?

• A)4. B)8 C)16 D)20 E)30.

• 2) Determine the total resistance of the circuit, if the resistance of the supply wires is 2 Ohms, the included part of the rheostat
• 64 Ohm and 294 Ohm lamps (Fig. 159).

• 1. 240 Ohm; 2. 180 Ohm; 3. 100 Ohm; 4. 120 Ohm; 5. 360 Ohm.

• 3) When measuring the voltage on conductor R 1, it turned out to be equal to 12 V. When the voltmeter was connected to conductor R 2 , then it showed 45 V (Fig. 160). Calculate the resistance R 2 if R 1 = 40 Ohms.

• A)360 Ohm; B)135 Ohm; B)150 Ohm; D)4 Ohm; D)40 Ohm.

• 4) In each of the two heating elements of the boiler, the current strength is 5 A. Determine the current strength in the supply wires if the elements are connected in series.

• A)25 A; B)5 A; B)10 A; D) 2.5 A.

• 5) Conductors with a resistance of 2.4 and 6 Ohms are connected in series and connected to a network with a voltage of 36 V. Calculate the current strength in the conductors.

• A)3 A; B)0.33 A; B)432 A; D) 0.5 A; D) 0.3 A.

• 1) Current strength in the conductor R 1 is equal to 4 A. What is the current strength in the conductor R 2 (Fig. 161).

• A)4 A; B) 2 A; B) 8 A; D)16 A.

• 2) The resistance of the lamp is R 1 = 300 Ohms, and the voltage across it is 90 V. What will the voltmeter show if it is connected to a lamp with a resistance of R2 = 400 Ohms (Fig. 162)?

• A)240 V; B)180 V; B)100 V; D)120 V; D)360 V.

• 3) Three identical lamps are connected in series to a 120 V network (Fig. 163). What is the voltage on each of them?

• A)360 V; B)120 V; B)60 V; D)4 V; D)40 V.

• 4) Figure 164 shows a stepped rheostat in which resistance R 1= R 2= R 3=…= R 5=10 Ohm. Calculate the resistance at a given position of the moving contact K.

• A)20 Ohm; B)50 Ohm; B)40 Ohm; D) 30 Ohm; D) 3.3 Ohm.

• 5) Electric lamp resistance R and the ammeter was connected to a 200 V network as shown in Figure 165. Calculate the resistance R , if the ammeter shows a current of 0.5 A. The lamp resistance is 240 Ohms.

• A)120 Ohm; B)160 Ohm; B)260 Ohm; D) 60 Ohm.

• A resistor with a resistance of 2 (Ohm) is connected to a circuit with a voltage of 12V. What resistance should another resistor be connected so that the current is 2A?

Repetition: series connection of conductors

• Two resistors and a light bulb are connected in a circuit with a source voltage of 12V. The voltage on the light bulb is 5V, on the first resistor 3V. The resistance of the second resistor is 6 (Ohm). Determine the resistance of the first resistor and the light bulb

• The current strength in the unbranched part of the circuit is equal to the sum of the currents in the branches
• The voltage on all parallel sections is the same
• The reciprocal value of the total resistance is equal to the sum of the reciprocal values ​​of the resistances of all parallel sections

Tasks for parallel connection of consumers

The resistor resistances are respectively 4,6,12(Ohm). Determine the current in each resistor if the voltage between points A and B is 24V. Determine the current strength in the unbranched part of the circuit

The current strengths in the resistors are respectively 2A, 1.5A, 3A. Determine the resistance of the resistors if the voltage between points A and B is 16V.

D/z § 48,49 exercise 22(1,2), exercise 23(3)

Slide 2

Series connection of conductors

When connecting conductors in series, the end of one conductor is connected to the beginning of another, etc. The pictures show a circuit connecting two light bulbs in series and a diagram of such a connection. If one of the bulbs burns out, the circuit will open and the other bulb will go out.

Slide 3

Laws of serial connection

When the conductors are connected in series, the current strength in all sections of the circuit is the same: According to Ohm's law, the voltages U1 and U2 on the conductors are equal: The total voltage U on both conductors is equal to the sum of the voltages U1 and U2: where R is the electrical resistance of the entire circuit. It follows: With a series connection, the total resistance of the circuit is equal to the sum of the resistances of the individual conductors.

Slide 4

Parallel connection of conductors

When conductors are connected in parallel, their beginnings and ends have common connection points to the current source.

Slide 5

Laws for parallel connection of conductors

With a parallel connection, the voltages U1 and U2 are the same in all sections of the circuit: The sum of the currents I1 and I2 flowing through both conductors is equal to the current in the unbranched circuit: Writing on the basis of Ohm’s law: where R is the electrical resistance of the entire circuit, we obtain When the conductors are connected in parallel, the value , the reciprocal of the total resistance of the circuit, is equal to the sum of the reciprocal values ​​of the resistances of parallel-connected conductors.

Slide 6

Problem 1Two conductors are connected in series. The resistance of one conductor is R = 2 Ohms, the other is R = 3 Ohms. The reading of the ammeter connected to the first conductor is I = 0.5 Ohm. Determine the current flowing through the second conductor, the total current in the circuit, and the total voltage in the circuit.

Slide 7

The solution of the problem

Given: R1= 2 OhmR2= 3 OhmI1= 0.5 AR Solution: I1= I2= Iu; I2= Iu= 0.5 AU1= I1R1; U1= 0.5 x 2= 1 (V)U2= I2R2; U2= 0.5 x 3= 1.5 (V)Uu= U1+U2; Uu= 1+1, 5 = 2.5 (V) I2, Iu, Uu=? Answer: I2= Iu= 0.5 A, Uu= 2.5 V.

Slide 8

Task 2.