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