Finding total resistance in parallel circuit

[kriegera]

Homework Statement

This is a 3-part problem of which I have solved 1 and 2. I've included here just so you can follow the problem. A 50-watt lamp, a 20-watt, a 75 watt lamp, and a 100-watt lamp are connected in parallel to a 110-volt circuit. Calculate a) the current through each lamp; b) the resistance of each lamp; c) the total resistance of the circuit.

Homework Equations

The Attempt at a Solution

a.) Current is calculate by I=P/V where P is in Watts so 
I=50/110 = .45 A
I=20/110 = .18 A
I=75/110 = .682 A
I=100/110=.91 A
b.) Here we use Ohm’s Law again: R = V / I
For 50 Watt lamp  R=110/.45 = 244.4 Ω
For 20 Watt lamp  R=110/.18 = 611 Ω
For 75 Watt lamp  R=110/.682 = 161.29 Ω
For 100 Watt lamp  110/.91 = 120.88 Ω


c.)To find total resistance of the circuit in parallel, use
R=1/(1/244.4) + (1/611) + (1/161.29) + (1/120.88) = 49.5 Ohms
I've seen many ways to calculate resistance - but is it accurate to do it this way for a parallel circuit?

 

[collinsmark]

c.)To find total resistance of the circuit in parallel, use
R=1/(1/244.4) + (1/611) + (1/161.29) + (1/120.88) = 49.5 Ohms
I've seen many ways to calculate resistance - but is it accurate to do it this way for a parallel circuit?

Yes, that's the method to do it.

I assume that by your equation you mean

$$R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}}$$

However, you should probably go back and re-do your answers for part a and b. The precision you used (i.e. significant figures) was small for part a, then you used a higher precision for part b, based on your low-precision answers from part a. Doing so causes some of the significant figures in your answers to part b to be pretty meaningless.

 

[kriegera]

Yes, that's the equation I mean. I'm a little confused - what do you mean by "significant figures" were small and "higher precision"? thanks!

 

[collinsmark]

Yes, that's the equation I mean. I'm a little confused - what do you mean by "significant figures" were small and "higher precision"? thanks!

As an example of the work you did already, you had (for the 50 W bulb)

I=50/110 = .45 A

[...]

For 50 Watt lamp  R=110/.45 = 244.4 Ω

Notice, you've calculated out the resistance to 4 significant digits, but the current was only calculated out to 2 significant digits.

You might be more careful with significant digits, especially early on. Continuing this example, suppose we used 5 significant figures (digits) from the beginning. Then,

I=50/110= 0.45454 A

And R = 110/0.45454 = 242.00 Ω

 

[willem2]

It's easier to calculate the resistance by calculating $R = V/I_{total}$

 

[kriegera]

got it:
For 50 Watt lamp  R=110/.4545 = 242.0 Ω
For 20 Watt lamp  R=110/.1818 = 605.0 Ω
For 75 Watt lamp  R=110/.6818 = 161.3 Ω
For 100 Watt lamp  110/.9090 = 121.0 Ω
c.) To find total resistance of the circuit in parallel, use

R=1/(1/242.0) + (1/605.0) + (1/161.3) + (1/121.0) = 49.38 Ohms
THANKS!