In the circuit shown at right,
X, Y and Z represent three
light bulbs, each rated at
60 watts, 120 volts.
Assume that the resistances
of the bulbs are constant and do not depend on the current.
A) What is the resistance of each bulb?
B) What is the equivalent resistance of the three light
bulbs when arranged as shown?
C) What is the total power dissipation of this combination
when connected to a 120-volt source as shown?
D) What is the current in bulb X?
E) What is the potential difference across bulb X?
F) What is the potential difference across bulb Z ?
A) P = IV 60W = I (120V)
I = 0.5 A
V = IR
120 V = 0.5 A(R)
R = 240 Ω
B) Ryz = (240-1 + 240-1)-1 = 120 Ω
Rxyx = 120 Ω + 240 Ω = 360 Ω
C) P = V2/R = 120 V2 / 360 Ω = 40 W
D) V = IR
120V = I(360 Ω)
I = 0.33 A
This is IT , but since X is in series with the other two it
receives the total current.
E) Vx = Ix Rx
Vx = 0.33A(240 Ω) = 80 V
F) VT = VX + Vyz
120V = 80V + Vyz
Vyz = 40V = Vy = Vz
The temperature at state 1 is 373 K. The gas goes through the
reversible cycle 1 ---> 2 ---> 3.
A) Determine the temperatures at state 2 and state 3.
B) Calculate the net work done on the gas.
C) Calculate the heat taken in by the gas.
A) V1/T1 = V2/T2 0.25m3/373K = 0.5m3/T2 T2 =746 K
B) P1/T1 = P3/T3
1.0x105/373 = 1.5x105/T3 T3 = 560 K
C) Work is done by the gas when it expands. Work is done
on the gas when it is compressed. Work is the area under
the line/curve. The gas does work from 1 --> 2 and work is
done on the gas from 2 --> 3. The difference between
these is the area in the triangle.
A = ½bh = .5(.25 m3)0.5x105 = 6250 J (This is done on the
gas since the compression part is larger.
D) ΔU = Q + W
U = Internal energy
Q = Heat added to system
W = Work done on the system
In a complete cycle, ΔU = 0.
Therefore Q = -W = - 6250 J (Heat is removed.)