How Does Current Flow Differ in Parallel vs Series Light Bulb Circuits?

[lampshader]

Homework Statement

If you were to make a circuit with two 150-Watt light bulbs in parallel -- how much current would flow through each bulb? Compare that to the amount of current which would flow through the bulbs if they were connected in series. Which way would create more light?

Homework Equations

1/R (base eq) = 1 / R (base1) + 1 / R (base2) + 1 / R(Base N) ...
I = E (base eq) / R (base eq)
P (base w) = I^2*R

The Attempt at a Solution

We can find the total resistance by:

1/R (base eq) = 1 / R (base1) + 1 / R (base2) = 1/96 Ohms + 1/96 Ohms
= 1/48 Ohms = 0.2080 Ohms^-1
= R (base eq) = 1 / 0.2080 Ohms^-1 = 4.8077 Ohms.

We can now find the current by using:

E (base eq) = 12V
R (base eq) = 4.8077 Ohms;

I = E (base eq) / R (base eq) = 12V / 4.8077 Ohms. = 2.496 =~ 2.5 A

P (base w) = I^2*R = (2.5 A)^2 * (4.8077 Ohms) = 30.0481 W

Therefore, 30.0481 Watt’s are flowing through each bulb during a live parallel circuit.

To find the current of two 150 Watt light bulbs in a series we can use:

R (base eq) = 96 Ohms + 96 Ohms = 192 Ohms.

and:

I = E (base eq) / R (base eq) = 12V / 192 Ohms = 0.625 A
So, P (base w) = I^2*R = (0.625 A)^2 * 192 Ohms = 0.75 W.


My Conclusion, therefore, is the parallel circuit with two light bulbs has more current than that of the series circuit.

Just wondering if I have done this problem correctly..

 

[willem2]

= 1/48 Ohms = 0.2080 Ohms^-1

You missed a 0

I = E (base eq) / R (base eq) = 12V / 192 Ohms = 0.625 A

here too.

So, P (base w) = I^2*R = (0.625 A)^2 * 192 Ohms = 0.75 W.

but now you end up with the right answer for the series case.


BTW Nothing in the problem says what voltage is the power source used.
IF they are 110V bulbs I'd connect the parallel circuit to 110V and the series
one to 220V and you'd get 300W total in each case.

 

[lampshader]

We can find the total resistance by:
1/R (base eq) = 1 / R (base1) + 1 / R (base2) = 1/96 Ohms + 1/96 Ohms
= 1/48 Ohms = 0.02080 Ohms^-1
= R (base eq) = 1 / 0.02080 Ohms^-1 = 48.077 Ohms.

We can now find the current by using:
E (base eq) = 12V
R (base eq) = 48.077 Ohms;

I = E (base eq) / R (base eq) = 12V / 48.077 Ohms. = 0.2496 =~ 0.25 A
P (base w) = I^2*R = (0.25 A)^2 * (48.077 Ohms) = 3.00481 =~ 3.005 W

Therefore, 3.005 Watt’s are flowing through each bulb during a live parallel circuit.

To find the current of two 150 Watt light bulbs in a series we can use:
R (base eq) = 96 Ohms + 96 Ohms = 192 Ohms.
and:
I = E (base eq) / R (base eq) = 12V / 192 Ohms = 0.625 A
So, P (base w) = I^2*R = (0.625 A)^2 * 192 Ohms = 0.75 W.

My Conclusion, therefore, is the parallel circuit with two light bulbs has more current than that of the series circuit.