Ohm's Law Problems: Solving for Resistance and Current

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SUMMARY

This discussion focuses on solving Ohm's Law problems related to resistance and current. Key calculations include determining the potential difference across a 220-Ω resistor with a current of 3.50 A, which yields a voltage of 770 V, and calculating the resistance of a 40-W lamp drawing 0.33 A at 120 V, resulting in a resistance of 363.6 Ω. Additionally, the operating resistance of a light bulb in a three-cell flashlight is calculated to be 5 Ω, and the analysis confirms that the device obeys Ohm's Law with a consistent resistance of 50 Ω across varying voltages. The discussion also clarifies the roles of ideal ammeters and voltmeters in circuit measurements.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of electrical power calculations (P=I^2R)
  • Familiarity with series and parallel circuit configurations
  • Concept of ideal ammeters and voltmeters
NEXT STEPS
  • Study advanced applications of Ohm's Law in complex circuits
  • Learn about Kirchhoff's laws for circuit analysis
  • Explore the characteristics of non-ideal ammeters and voltmeters
  • Investigate power dissipation in resistive components
USEFUL FOR

Students, electrical engineers, and hobbyists interested in circuit analysis and practical applications of Ohm's Law.

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ohm's law problems. help...

Research Work:

What is an ideal ammeter and an ideal voltmeter? Why are voltmeters connected in parallel? Why are ammeters connected in series?

Problems:

answer any number that you know. thanks so much. i just neede it badly...

1.What is the potential difference across a 220-Ω resistor when a current of 3.50 A flows through it?

2.A 40-W electric lamp draws a current of 0.33 A when operated with a potential difference of 120 V. What is the resistance of the lamp?

3.A three-cell flashlight draws a current of 0.60 A. What is the operating resistance of the light bulb if each cell provides a potential of 1.0 V when delivering this current?

4.The current through an electronic device is measured for several different voltages applied across the device. When the potential difference is 0 V, 0.50 V, and 0.75 V, the current is 0 A, 0.010 A, and 0.015 A, respectively. Does the device obey Ohm’s Law?
 
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An ideal ammeter would be one that could measure the current in a circuit without changing it. It would have no resistance. An ideal voltmeter would be one that could measure the voltage difference between two points in a circuit without changing that voltage. It would have an infinite resistance.

Voltmeters measure the votage difference between two points in a circuit. If you connrcted one in series with a circuit, think about what you would be measuring. That should explain why they are generally connected in parallel.

Ammeters measure the current flowing through them and have very little resistance. Think about what would happen if you connected one in parallel. Since it would alter the current flowing in the rest of the circuit, what you would be measuring wouldn't be very meaningful.

1. Try Ohm's law: V=IR
2. Use the formula for the power, P, to solve for R: P=I^2R[/tex] (The problem gives extra information, but is not inconsistant with itself.)<br /> 3. Presumably the cells are connected in parallel and the total voltage is the sum of the ndividual voltages. From there it&#039;s just Ohm&#039;s law again.<br /> 4. Basically, just solve for R in Ohm&#039;s law for each voltage (other than 0, as this will be indeterminant). Is R the same in each case?
 
1/ V= 220*3.50 =770
2/ R = 120/0.33 =363.6
3/ R = (1+1+1)/0.6 =5
4/ V= 0*R, V=0.5=0.010*R => R =50, V=0.75=0.015*R => R=50 So YES.
 

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