The Physical World I

Spring 2005
MWF 11:30

Instructor: Dr. Bruce Law, CW 327, Tel: 532-1618.


R23.11. I = V/R, therefore if R doubles I must decrease by a factor of 2.

R23.12. I = V/R, if V is halved than I must also be halved.

R23.27. You are paying for the energy EPE = IVt, which depends upon the time t during which you are using the electrical equipment.

R23.29. Electric power = current x voltage.

R23.33. The current through the other lamp must be 1A because, in a series circuit, the same current flows through both lamps.

R23.35. The main shortcoming of a series circuit is that if one lamp blows then all the lamps go off because there is no longer any current flowing in the circuit.

R23.36. 6V is also applied across the other lamp in a parallel circuit.



E23.4. The brightness increases because more electrical energy is being converted to light (and heat).

E23.7. Circuit 5.

E23.8. The same amount of current flows out of a battery or light bulb as into it. Recall that the current is the number of electrons flowing through a circuit per second -- electrons do not accummulate in any part of the circuit, therefore, the outflow must be equal to the inflow.

E23.25. Automobile headlights must be wired in parallel because when one lamp blows the other lamp does not go out.

E23.40. The lightbulbs are identical and therefore have the same resistance. Lightbulb C draws the most current, this is because current I=V/R -- the resistance R is twice as large in line AB (because there are two light bulbs)compared with C, hence, the current in A and B is half the size as the current in C. If bulb A is unscrewed, bulb B will go out and C will remain unchanged. If C is unscrewed, A and B will remain unchanged.

E23.41. In a series circuit, as more bulbs are added, the brightness of each bulb decreases. This is because the resistance R increases which causes the current I through each bulb to decrease (recall I = V/R). In a parallel circuit, as more bulbs are added, the brightness of each bulb remains unchanged – each bulb acts independently of the other bulbs in this case.


 PROBLEMS OF Home Work 20

P23.2. Resistance = voltage/current = 120/20 = 6 ohms.

P23.3. Current = power/voltage = 1200/120 = 10A. Resistance = voltage/current = 120/10 = 12 ohms.

P23.7. Heat generated from EPE = IVt = 9x110x60 = 59400 J where time t = 60 seconds.

P23.8. Current = charge/time, therefore, charge = current x time = 9A x 60s = 540C.