How can I determine the relative brightness of lightbulbs in different circuits?

Question

Identical light bulbs are connected to identical batteries in the various circuits below. Use your model and previous observations to rank all the bulbs (A-H) in order of brightness (from greatest to least using > and = symbols, do alphabetize those that are equal). Explain your reasoning.

*Can't attach pictures, so I'll describe the circuits*

1) Battery and bulb A (single loop)

2) Battery and bulbs B and C connected in series (single loop)

3) Battery and bulbs D and E connected in parallel (two loops)

4) Battery and bulbs G and H connected in series, that combination in parallel with bulb F (two loops)

Answer 1

Answer 2

Method 1 - calculation

Use P=V²/R, where V is the voltage across the bulb, R is the impedance of each bulb. We will just focus on V since R is identical.

Denote battery voltage as U, battery impendence as r. It is reasonable to think R>>r, meaning R/2 should also be significantly larger than r.

Case1:) for A: V/U = R/(R+r) (highest)

Case 3: Combined D/E in parallel generates R/2 to compete with r for the battery voltage,

V/U = (R/2)/(R/2+r) = R/(R+2r) , worse than A. A>D=E

Case 4): Combined G/H/F impedance now becomes 2R/3.

For F: V/U = (2R/3)/(2R/3+r) = 2R/(2R+3r) , easy to see this is greater than case 3 D/E. (You can make the ratio to get V_F/V_DE = (2R+4r)/(2R+3r)>1). So F>D=E

For G and H, they are exactly half of F. so for them, V/U=R/(2R+3r), much less than D/E. So F>D=E>G=H

Case 2: combined B and C has impedance of 2R, so for B or C, V/U= R/(2R+r). This is greater than G/H but less than F. So we have A>F>D=E>B=C>G=H

Method 2: Intuitively

A is the brightest since it is the only one that competes with r, no one better.

D and E is slightly less than A, because the combined parallel impedance becomes R/2, but still >>r, they still take a significant amount of battery voltage.

F is better than D, because adding additional bulb will make the group take more battery voltage than D/E. But G/H will be really dim because it is only half of the F.

B/C is slightly half of the battery voltage, it will be less than D/E. Because remember D/E is only slightly less than A and B/C is about half of A.

What about B/C and G/H?

B/C is about half of A, and G/H is half of F. And F is about 2/3 of A due to combined impedance of F/G/H. (I guess you do need a little bit of calculation). So B/C is better than G/H

Same conclusion: A>F>D=E>B=C>G=H

Answer 3

Great question. A good place to start would be with what we know already. First, they are identical batteries, which means each of the circuits will have the same voltage drop from one end of the battery to the other. Second, they are identical light bulbs. This means the resistance R of each individual light bulb is the same.

The power dissipated by each of the light bulbs, which is directly related to the brightness, is given by the formula P=I² R. Since all of the light bulbs have the same resistance, we just need to find the current I going through each of the bulbs.

For the first circuit A, this can be found using Ohm's Law since there is only one bulb in the system. V=IR -> I=V/R where V is the voltage of the battery.

For the second circuit, there are two light bulbs in series. Resistances in a circuit add when they are in series with each other. This means the resistance of the whole circuit will be 2R. Using this with Ohm's Law, we find the current going through each of the bulbs will be I=V/2R for bulbs B and C.

For the third circuit (D & E), the two bulbs are connected in parallel. When resistances are in parallel, their inverses add. To find the resistance of the whole circuit, we perform the addition of the inverses of resistances like : (1/R_circuit = (1/R) + (1/R) --> R_circuit=R/2). From this point, we know the only voltage drops in the circuit are coming from the parallel light bulbs. This means, no matter which path is taken, the voltage drops by the voltage V of the battery. Using Ohms Law and the resistance of the light bulb, we can find that the current through either of the light bulbs is I=V/R (the same as A) for both D & E.

For the 4th circuit, we will combine the concepts from the last two situations. First, we want to find the equivalent resistance of the two light bulbs in series on one side of the parallel loop. There are two bulbs in series, so we can add their resistances to form a equivalent resistance for that side of the parallel loop as 2R. The resistance of the other side is R. Since we know that either way the current goes, the voltage drop must be the same as the battery since there are no other components in the circuit. For the sole light bulb side F, this current is I=V/R (same as A). For the other side of the parallel portion, the current flowing through both bulbs is I=V/(R+R) = V/2R. This will be the current for both G and H.

Since the power and brightness are directly related to the current, and each light bulb has the same resistance, the order should be:

A=D=E=F>G=H=B=C

I hope this helps!