In the circuit below, what wake up to the brightness that the two light bulbs as soon as the move is closed, connecting the third light pear to the circuit? You deserve to assume that light bulbs are like resistors, and also that a bulb"s brightness is proportional come the existing through it.

You are watching: Does bulb a get brighter, dimmer, or stay the same brightness? $$a.$$ Bulbs 1 and 2 gain brighter$$b.$$ Bulbs 1 and also 2 obtain dimmer$$c.$$ bulb 1 it s okay brighter and bulb 2 gets dimmer$$d.$$ bulb 1 gets dimmer and also bulb 2 gets brighter$$e.$$ Bulbs 1 and also 2 continue to be the same This is a theoretical problem, us don"t have any kind of numbers. We space told, however, the the bulbs act prefer resistors, and also that brightness is proportional come the current. This way we can redraw ours figure and start investigate what will happen.  We want to know about current, however we are offered voltage (the source powering the bulbs). For this reason we"re not sure if VDR or CDR will assist us simply yet. What has adjusted between these 2 states? We put in another resitor, so there was definitely a readjust in indistinguishable resistance. We included another aspect in parallel v resistor 2. And we know that including an facet in parallel lowers resistance.  Since resistance to be lowered, however voltage continued to be the same, that method current has to rise come compensate.the complete current with the whole resistor network is same to the current through facet 1, since element 1 is in collection with the remainder of the circuit.The present in aspect 1 rises, due to the fact that the total current rose.Bulb 1 it s okay brighterThis eliminates answer B, D, and also E  Now we have to number out what happens to pear 2 without understanding anything around the brand-new resistance exept the it is positiveWe uncovered out in the previous section that aspect 1 has much more current. This means, through Ohm"s law, the it is also consuming much more voltage.If aspect 1 is consuming more voltage, then that method that the rest of the circuit (element 2 and the new element) must have actually a lower voltage drop than before, because the two must add up come the resource in either case.And resistor 2 is in a parallel configuration. The voltage across a parallel configuration is identical to the voltage throughout each of the elements. That voltage simply dropped.And with reduced voltage comes reduced current, and with lower present comes reduced brightness.
Therefore, bulb 2 gets dimmer
For those unsatisfied through the conceptual analysis this is the full mathematical proof for the trouble above.
\<\beginalign&R_open&=&R_1+R_2&\\ \\&I_open R_open&=&V_B&\\\\&I_open&=&\fracV_BR_open&\\\\&I_open&=&\fracV_BR_1+R_2&\\\\&I_1,open&=&\fracV_BR_1+R_2&\\\\&I_2,open&=&\fracV_BR_1+R_2&\endalign\>
First, let"s completely anayze the initial, open switch stateI"ll usage $$R_open$$ to suggest the total equivalent resistance when the switch is open. $$R_1$$ and also $$R_2$$ room the resistors the the two bulbs $$V_B$$ is the voltage that the resource (presumably a battery)$$I_open$$ is the current through the battery when the switch is open.$$I_1,open$$is the present through pear 1 when the move is openSince $$R_1$$ and $$R_2$$ room in series, castle both have currents equal to the battery current.
\<\beginalign&R_closed&=&R_1+\fracR_2 R_3R_2+R_3&\\\\&I_closed R_closed&=&V_B&\\\\&I_closed &=& \fracV_BR_closed&\\\\&I_closed &=& \fracV_BR_1+\fracR_2 R_3R_2+R_3&\\\\&I_1,closed&=&I_closed=\fracV_BR_1+\fracR_2 R_3R_2+R_3&\\\\&I_2,closed&=&I_1,closed \fracR_3R_3+R_2&\\\\& &=&\fracV_BR_1+\fracR_2 R_3R_2+R_3 \fracR_3R_3+R_2&\endalign\>
Now we do the same evaluation on the closeup of the door switch. Though it was unlabeled in the original problem, I will certainly label the brand-new resistor $$R_3$$.The existing coming the end of the battery currently uses the brand-new resistance. $$I_1$$ matches this, however $$I_2$$ go not, since there is currently an extra course or the current. The present Divider dominion gives united state the separation of $$I_closed$$ in between $$R_2$$ and also $$R_3$$
Now the we have actually expressions, we have the right to compare. An initial we compare $$I_1,open$$ to $$I_1,closed$$.If your psychological math is strong, girlfriend can notification that the denominator in the expression because that $$I_1,closed$$ is definately much less than the one because that $$I_1,open$$, and division by a smaller quantity gives a larger finish result.
\<\beginalign&I_1,closed&=&\fracV_BR_1+\fracR_2 R_3R_2+R_3&\\&I_1,closed&=&\fracV_BR_1+\fracR_2 R_3R_2+R_3 \fracR_1+R_2R_1+R_2&\\&I_1,closed&=&\fracV_BR_1+R_2 \fracR_1+R_2R_1+\fracR_2 R_3R_2+R_3&\\&I_1,closed&=&I_1,open\fracR_1+R_2R_1+\fracR_2 R_3R_2+R_3&\\&I_1,closed&=&I_1,open\fracR_1+R_2R_1+\fracR_2 R_3R_2+R_3 \fracR_2+R_3R_2+R_3&\\&I_1,closed&=&I_1,open\frac(R_1+R_2)(R_2+R_3)R_1(R_2+R_3)+R_2 R_3 &\\&I_1,closed&=&I_1,open\underbrace\frac(R_1 R_2+R_2 R_3 +R_3 R_1) +R_2^2(R_1 R_2+R_1 R_3+R_2 R_3)_>1&\endalign\>
If that discussion isn"t convincing, we can rewrite the expression for I_1,closed to do it have actually a much more obvious connection to $$I_1,open$$We can present $$I_1,closed$$ to be equal to $$I_1,open$$ time some huge factor, and that aspect is definitely bigger than 1. Two points on top, among them top top bottom. Bulb 1 becomes brighter after the move is closed

$$I_2,open=\fracV_BR_1+R_2$$Vs$$I_2,closed=\fracV_BR_1+\fracR_2 R_3R_2+R_3 \fracR_3R_3+R_2$$
\<\beginalign&I_2,closed&=&\fracV_BR_1+\fracR_2 R_3R_2+R_3 \fracR_3R_3+R_2&\\&I_2,closed&=&\fracV_BR_1+R_2 \fracR_1+R_2R_1+\fracR_2 R_3R_2+R_3 \fracR_3R_3+R_2&\\&I_2,closed&=&\fracV_BR_1+R_2 \fracR_1+R_2R_1 (R_2+R_3)+R_2 R_3 \fracR_31&\\&I_2,closed&=&I_2,open\fracR_3 R_1+R_3 R_2R_1 R_2+R_1 R_3+R_2 R_3 &\\&I_2,closed&=&I_2,open \underbrace\frac(R_3 R_1+R_3 R_2)R_1 R_2+(R_1 R_3+R_2 R_3) _{To execute this, we just cheat. Us multiply through something equal to 1, i m sorry gies us the denominator we desire times a bunch of junkDistruibute the (R2+R3) termDistribute moreWe now have the original expression times a quantity that is absolutely less than 1, since it is something divided by itself added to an additional positive number pear 2 gets dimmer.

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As deserve to be seen, the qualitative way of fixing things can gain us a little result, favor "increases" or "decreases" for reasonably little effort. Come get specifically HOW much it increases, we have to do some math. And also by part math I typical a most tedious algebra.