51) This question asks us to find the circuit which will have the brightest bulbs. The brightness of the bulbs is determined by the power dissipated by the bulb.
Here's a simple way we can solve this problem. All of the circuits have at least 1 series connection except for circuit 2. This is important because it means there is a lower voltage drop when the bulbs are connected in series. Thus circuit 2 would have the brightest bulbs.
Answer: B
NOTE: This method only works for this problem because all lightbulbs are identical and 1 of the circuits has all the bulbs in series. These types of solutions are like estimates and may not work for every problem. The rest of the explanation for this problem going into depth about getting exact answers.
We will use the red equation because we don’t need to involve current(I) which would be harder to solve for.
The problem tells us that all the lightbulbs have equivalent resistance, so let’s go through each of the circuits.
Circuit 1: This circuit has three lightbulbs all in series. Since they are all in series and have the same resistance, there will be the same voltage drop across each resistor. And the sum of the voltage drops should sum to the source voltage(V). Since they have the same voltage and same resistance, all three resistors will have the same power dissipated.
All three resistors will dissipate (V^2)/(9R) watts.
Circuit 2: This circuit has 3 resistors all in parallel to each other. Since there are no resistors in series, the voltage drop between any parallel will be equal to the total source voltage.
All three resistors will dissipate (V^2)/R watts.
Circuit 3: This circuit has a lightbulb connected in series with 2 bulbs connected in parallel. Our first step is to find the equivalent resistance.
Since R[1] is 2/3 of the equivalent resistance, the voltage drop across R[1] will be 2/3 of the source voltage. Likewise, the voltage drop across R[2] and R[3] will be 1/3 of the source voltage.
Now that we have the voltage drop across each light bulb, we can find the power dissipated by each bulb.
Circuit 4: This circuit gives us a parallel circuit where two lightbulbs connected in series are connected parallel to a single lightbulb.
The voltage drop between each parallel will be the same; however, on the top parallel, the voltage drop is split among the two light bulbs on that parallel.
As shown above, R[1] and R[2] are each half of the resistance across the parallel. This means we can find the voltage drop from each resistor.
Finally, we can solve for the power dissipated in each bulb in this situation.
From this, we can conclude that the lightbulbs all shine brightest in situation 2.
Answer: B
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