{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ohm - Ohm's Law In many materials the voltage and...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Ohm's Law In many materials, the voltage and resistance are connected by Ohm's Law: Ohm's Law : V = IR The connection between voltage and resistance can be more complicated in some materials.These materials are called non-ohmic. We'll focus mainly on ohmic materials for now, those obeying Ohm's Law. Example A copper wire has a length of 160 m and a diameter of 1.00 mm. If the wire is connected to a 1.5-volt battery, how much current flows through the wire? The current can be found from Ohm's Law, V = IR. The V is the battery voltage, so if R can be determined then the current can be calculated. The first step, then, is to find the resistance of the wire: L is the length, 1.60 m. The resistivity can be found from the table on page 535 in the textbook. The area is the cross-sectional area of the wire. This can be calculated using: The resistance of the wire is then: The current can now be found from Ohm's Law: I = V / R = 1.5 / 3.5 = 0.428 A Electric power Power is the rate at which work is done. It has units of Watts. 1 W = 1 J/s
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern