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Two circuits are created using two identical light bulbs. In the circuit A, the bulbs are hooked up in series. In circuit B, the light bulbs are hooked up parallel. Each circuit is powered by a 6-volt battery.
Which statement or observation would not be supported by the two circuits?
a) The current in the circuit A would be less.
b) The light bulbs in the circuit B would be brighter
c) The light bulbs in circuit B are equal in brightness.
d) If one light burns out in circuit A, the other bulbs gets brighter.
1 Answer
- electron1Lv 72 years ago
The easiest way to solve this problem is to let the resistance of each bulb be 0.5 Ω.
When the bulbs are connected in series, the total resistance is the sum of the resistances.
R = 1 Ω
I = 6 ÷ 1 = 6 amps
In parallel, 1/Req = 1/0.5 + 1/0.5 = 4 Ω
I = 6 ÷ 4 = 1.5 amps
The current in circuit A is the greatest. The brightness is dependent on the power.
Power = V * I
For series, P = 6 * 1 = 6 watts
For parallel, P = 6 * 1.5 = 9 watts
The power is greater in the circuit B.
d) If one light burns out in circuit A, the other bulbs gets brighter.
Since the bulbs are in series, when one light bulb burns out, the current will stop flowing. This will cause both lights to go out. If you did this is circuit A, the other bulb would get brighter. I hope this is helpful for you.
