ch27-p007

ch27-p007 - the rate P = i E if the current and emf are in...

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

7 (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 .U s e Kirchhoff’s loop rule: E 1 iR 2 iR 1 −E 2 = 0. Solve for i : i = E 1 −E 2 R 1 + R 2 = 12 V 6 . 0V 4 . 0 +8 . 0 =0 . 50 A . A positive value was obtained, so the current is counterclockwise around the circuit. (b) and (c) If i is the current in a resistor with resistance R , then the power dissipated by that resistor is given by P = i 2 R .Fo r R 1 the power dissipated is P 1 =(0 . 50 A) 2 (4 . 0 )=1 . 0W and for R 2 the power dissipated is P 2 =(0 . 50 A) 2 (8 . 0 )=2 . 0W . (d) and (e) 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: the rate P = i E if the current and emf are in opposite directions. For battery 1 the power is P 1 = (0 . 50 A)(12 V) = 6 . 0 W and for battery 2 it is P 2 = (0 . 50 A)(6 . 0 V) = 3 . 0 W . (f) and (g) In battery 1, the current is in the same direction as the emf so 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

{[ snackBarMessage ]}

Ask a homework question - tutors are online