Physics Question?

Question

I can't seem to come up with the right combination of equations to solve this problem. I know that V=IR and P=IV, but I'm not having any luck with the substitutions I'm making. Can anyone give me some guidance on how to approach a problem like this and find two different resistance values? Not looking for the answer, as I need to solve it myself, but I am in desperate need of guidance. Any help would be greatly appreciated!

A light bulb is wired in series with a 132 ohm resistor, and they are connected across a 138.0 V source. The power delivered to the light bulb is 22.7 W. What are the two possible resistances of the light bulb?

low value= ?
high value=?

Answer 1

If you follow me.as, 's method, you should find X to be either 590 or 30 ohm, both approx.

You can also approach it like this.
1. If x is 0, ie, the bulb has no resistance, then, current thru the circuit is 138/132, almost 1 amp. In fact, if you guess x = 6 ohm, then total R = 138 ohm and i=1 am and power thru bulb is 1*1*6= 6 watt, which is a little low.
Now, if somehow x increases to 2*6+132 = 144, so that total resistance is 144+132= 276, then i = 0.5 amp, and power to bulb would be 0.5*0.5*276 = 69 watt, already high.

Play with it and you may arrive at a low value that works.

2. Approach it as the high side conceptually is that if the bulb has infinite resistance (open circuit), then there is no current and therefore no power consumption. So if the bulb starts out with a higher resistance than the resistor, the current is small and that makes i-square even smaller but you are multiplying it to a large resistance value. So if X is much larger than the resistor that you can just call the total resistance X, then power is V*V/total R

Make 138*138/R =22.7. you get total R = 839

You realize that 839 is not that much larger than 132. But you see you are closing in.

The key points are that you have no power consumption at the light bulb when its resistance is very large or very small. That is why there is a high and low value.

Answer 2

Let X be the resistance of the light bulb.

Now we know,V= IR
In this case V= 138 V
I = got to find out
R = 132 + X ( in a series circuit, the total resistance is the summation of the individual resistances)

In a series circuit equal current flows through all the resistors. So we got ot find the value of I.

P=IV and V=IR so substituting for V
P= I (IR) = I^2 R
In this case P = 22.7 ( for the light bulb)
R = X ( resistance of light bulb)
So I = sqrt(P/X) = sqrt(22.7/X)

Now, for the circuit,
V=IR
138 = sqrt(22.7/X) . (R+X)

Solve the quadratic equation to get the two values of resistances (X). But remeber resistances cant be negative. So if you get a negative value discard it.