Spring 2005

MWF 11:30

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

# REVIEW QUESTIONS OF Home Work 20

# 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.

#

# EXCERISES OF Home Work 20

## 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. **