8B-Soln07-v1.0.4

# 8B-Soln07-v1.0.4 - terminal(current is flowing clockwise...

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

terminal (current is flowing clockwise), and so consumes power. Both bulbs dissipate energy and so also consume power. Note that P batt1 = 60 W, P bulb1 = 36 W, P bulb2 = 4 W, and P batt2 = 20 W, so P batt1 - P bulb1 - P bulb2 - P batt2 = 0. That’s just conservation of energy - power delivered minus power dissipated equals zero. h) The positive charges will be stationary. The free electrons are moving to the right , since the current is pointing to the left. 6. R ext = 0.2 Ω R ext = 11 Ω a) I = 10 Amps I = 1 Amp b) V ext = 2.0 V V ext = 11 V c) P ideal batt = 120 Watts P ideal batt = 12 Watts d) P ext = 20 Watts P ext = 11 Watts e) Energy has been dissipated as heat as a result of current passing through the internal resistor in both cases. (100 Watts dissipated for the 0.2 Ω setup, and 1 Watt dissipated for the 11 Ω setup) The power delivered by the ideal battery is the same as the sum of power dissipated by the internal resistor and the external resistor. 7. a) i)
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/10/2012 for the course PHYSICS 8B taught by Professor Shapiro during the Spring '07 term at Berkeley.

Ask a homework question - tutors are online