P28_005

# P28_005 - P = i E if the current and emf are in opposite...

This preview shows page 1. Sign up to view the full content.

5. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 .W eu s e KirchhoF’s loop rule: E 1 iR 2 iR 1 −E 2 =0. Wesolvefor i : i = E 1 −E 2 R 1 + R 2 = 12V 6 . 0V 4 . 0Ω+8 . 0Ω =0 . 50 A . A positive value is obtained, so the current is counterclockwise around the circuit. (b) If i is the current in a resistor R , then the power dissipated by that resistor is given by P = i 2 R . ±or R 1 , P 1 =(0 . 50 A) 2 (4 . 0Ω)=1 . 0 W and for R 2 , P 2 =(0 . 50 A) 2 (8 . 0Ω)=2 . 0W. (c) If i is the current in a battery with emf E , then the battery supplies energy at the rate P =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P = i E if the current and emf are in opposite directions. ±or E 1 , P 1 = (0 . 50 A)(12 V) = 6 . 0 W and for E 2 , P 2 = (0 . 50 A)(6 . 0 V) = 3 . 0 W. In battery 1 the current is in the same direction as the emf. Therefore, this battery supplies energy to the circuit; the battery is discharging. The current in battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging....
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online