2004 phy al mc Q21

Notes that Vx+Vy must equal to 200V.
Also, Current should be the same.
In the figure, When I=0.25A,Vx=50,Vy=150,where 50+150=200V
=>It is a possible result.
By P=IV.Done!

You could only use approximation in this question.

Bulb X is of rated power 200v, 100w, whereas bulb Y is of rated power of 200v, 60w. By the formula, power P= V^2/R (V is the voltage and R is the resistance of the bulb), resistance of bulb X is smaller than that of bulb Y.

Hence, when the two bulbs are connected in series, because the current flowing through the two bulbs is the same, a higher resistant bulb will consume higher power. Thus, the power of bulb Y is higher than that of bulb X. Hence, option C and D must be wrong.

Now, the voltages across X and Y, when added together, should be equal to 200v. From the given curves, try to locate a current such that the sum of voltages of X and Y (i.e. the values at the x-axis) is 200v. This current is approximately 0.24 A, at which the voltage of X (the value at x-axis) is 50v, and the voltage of Y is 150v.

The estimated power of bulb X = 0.24 x 50 w = 12 w
The estimated power of bulb Y = 0.24 x 150 w = 36 w
.

2010-06-10 21:20:46 補充：
You could also use a simple trial calculation to help you test option A or B is the correct answer. Let V be the voltage across X, voltage across Y = (200-V). Hence, first try option A: VI = 12, (200-V).I = 36 (where I is the current)....

2010-06-10 21:27:41 補充：
(cont'd)...Dividing and solve for V gives V = 50v, voltage of Y is thus 150v. This is the right option as the current, as seen from the graph, is the same for both X and Y.

2010-06-10 21:27:59 補充：
When try option B, VI=15, (200-V)I=25, solve for V gives V = 75v, (200-V)=125v. Clearly, this is not the right option, current for the two bulbs, as seen from the two curves, are not the same. This contradict what current in a series conection should be.