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Order the circuits according to their power usage, from highest to lowest. All batteries are the same voltage, and all light bulbs have the same resistance. A battery connected to three light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the third bulb. The third bulb connects to the negative battery terminal. Circuit A A battery connected to two light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit B A battery connected to two light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the negative battery terminal. Circuit C A battery connected to three light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit D Highest power usage Lowest power usage

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Answer:

  P_a = 3P₀ ,  P_b = 2 P₀ ,       P_c = 3 P₀

P_c> P_b> P_a

Explanation:

To determine which circuit consumes more energy than is given by the expression

           P = V I

           V = I R

           I = V / R

           P = V² / R

As all circuits have the same battery, the value of the resistance to which the battery is connected determines the consumption

Circuit A

In this circuit the three bulbs are in series so the total resistance is

         R_total = 3 R

the power dissipated is

         P_a = V² / 3R

if we call

         P₀ = V² / R

       

we substitute

         P_a = P₀/3

Circuit B

Two bulbs are connected in series

           R_total = 2 R

power is

        P_b = V2 / 2R

        P_b =  P₀/2

Circuit C

The 3 bulbs are connected, but in parallel, the resistance is

        1 / R_totak = 1 / R + 1 / R + 1 / R

        R_total = R / 3

        P _c = V2 3 / R

        P_c = 3 Po

By reviewing these results, we can sort the circuits

        P_c> P_b> P_a

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