"Application of Ohm's Law in Series and Parallel Circuits" Ohm's Law PPT Courseware

Description

"Application of Ohm's Law in Series and Parallel Circuits" Ohm's Law PPT Courseware

review

1. Ohm's law

The current in a conductor is directly proportional to the voltage across the conductor and inversely proportional to the resistance of the conductor.

2. Current and voltage rules in series circuits

(1) The current in the series circuit is equal everywhere; I=I1=I2

(2) The total voltage in a series circuit is equal to the sum of the voltages at both ends of each part of the circuit. U=U1+U2

3. Current and voltage rules in parallel circuits

(1) The current in the trunk of the parallel circuit is equal to the sum of the currents in each branch; I=I1+I2

(2) The voltages at both ends of each branch in a parallel circuit are equal. U=U1=U2

1. Series and parallel connection of resistors

Want to discuss

Previously we have learned the laws of current and voltage in series and parallel circuits. So, when resistors are connected in series or parallel, is the total resistance larger or smaller than before?

Share your thoughts with your classmates and share your reasons.

The total resistance of the series resistors is greater than any of the partial resistors

The total resistance of a parallel resistor is smaller than any of its partial resistors

1. Series connection of resistors

From Ohm's law: U1=I1R1 U2=I2R2 U=IR

It can be seen from the series circuit: U =U1+U2 I=I1=I 2

So: IR=I1R1+I2R2

That is: R=R1+R2

Conclusion: The total resistance of a series circuit is equal to the sum of the individual resistances.

practice

1. There are two fixed value resistors A and B. A is marked with "16Ω 1.5A" and B is marked with "22Ω 0.5A". Now connect them in series and connect them to a circuit. The maximum voltage allowed to be applied to both ends of the circuit is ___V.

2. As shown in the figure, the resistance of R1 is 3Ω, the resistance of R2 is 6Ω, and the power supply voltage is 3 V. What is the current in the main circuit? (Please answer in two ways)

Example question analysis

Example 1 As shown in the figure, it is known that I=4.5 A, I2=0.9 A, and the power supply voltage is 36 V. How big is the resistance R1 of the lamp L1?

Analysis: ∵ Lamp L1 and lamp L2 are connected in parallel

∴ I = I1 + I2

I1 = I - I2

= 4.5 A -0.9 A

= 3.6 A

And U1 = U2 =U= 36 V

∴R1=U1/I1=36 V/3.6 A=10 Ω

2. Application of Ohm’s Law

Voltmeter

Want to discuss

(1) Why must the ammeter be connected in series in the circuit under test?

(2) Why can’t the measuring ammeter be directly connected to both ends of the power supply?

(1) The current in a series circuit is equal everywhere.

(2) The internal resistance of the ammeter is very small. When directly connected to both ends of the power supply, too much current will pass through, which will damage the ammeter.

Ammeter

Want to discuss

(1) Why must the voltmeter be connected in parallel at both ends of the electrical appliance under test?

(2) Why can a voltmeter directly measure the voltage of the power supply?

(1) The voltages at both ends of each branch in a parallel circuit are equal.

(2) The internal resistance of the voltmeter is very large. When directly connected to both ends of the power supply, the current passing through is very small and will not damage the voltmeter.

Class exercises

1. As shown in the figure, the power supply voltage remains unchanged. After the switch is closed, if P moves to the right, the changes in the voltmeter and current indications are ( )

A. The voltage indication number becomes larger and the current indication number becomes smaller.

B. The voltage indication number becomes larger and the current indication number becomes larger

C. The voltage indication number becomes smaller and the current indication number becomes larger.

D. The voltage indication number becomes smaller and the current indication number becomes smaller

2. In the circuit shown in the figure, the power supply voltage is 6V. After the switch S is closed, the lamp L does not light up. The voltage across resistor R measured with a voltmeter is 6V, which means: ( )

A. Both the lamp L and the resistor R are intact

B. Lamp L is intact but resistor R is broken.

C. Lamp L is broken and resistor R is intact

D. Both the lamp L and the resistor R are open circuit

3. Xiaodong did an experiment to measure the resistance of a light bulb. After connecting the circuit and closing the switch, he found that the light did not light up, the ammeter showed no value, and the voltage indicated was the power supply voltage. What are the possible causes of the failure?

Please write down three reasons why the light bulb does not light up.

(1) The voltmeter is connected in series in the circuit;

(2) The small light bulb is broken;

(3) The wire or terminal connecting the two ends of the small bulb is broken.

Class summary

Steps to solve circuit calculation problems

(1) Analyze the connection methods between resistors in each circuit state according to the meaning of the question, and draw the equivalent circuit diagram.

(2) By reviewing the questions, clarify the known conditions and unknown quantities given in the questions, and mark the symbols, values ​​and units of the known quantities and the symbols of the unknown quantities on the circuit diagram.

(3) Each step of the solution process must include three steps: writing the formula - substituting values ​​and units - obtaining the result.

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Update Time: 2024-07-29

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