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Sagot :
We should connect the three bulbs in parallel in order to allow at least one bulb to use maximum power.
We have a 20 V power source and three lightbulbs that have the same resistance.
We have to design a circuit that would allow at least one bulb to use maximum power.
What is the total resistance of the circuit if three resistances each of of resistance R(1), R(2) and R(3) are connected in parallel ?
For parallel combination of resistances, the total resistance will be -
[tex]\frac{1}{R_{T} } =\frac{1}{R(1)} +\frac{1}{R(2)} +\frac{1}{R(3)} \\[/tex]
We know that the power dissipated by the resistance is equal to -
P = V x I = [tex]I^{2} R[/tex] = [tex]\frac{V^{2} }{R}[/tex]
Let the three bulbs be B(1), B(2) and B(3) each with resistance ' R '. In parallel combination of the bulbs, the voltage across each bulb will be same as that of power source -
V [B(1)] = V [B(2)] = V [B(3)] = 20 V
Therefore, the power used by each bulb will be -
P [B(1)] = P [B(2)] = P [B(3)] = [tex]\frac{20\times 20}{R}=\frac{400}{R}[/tex]
Whereas, in series combination the voltage drop will regularly take place after the current passes through a resistor.
Hence, we should connect the three bulbs in parallel in order to allow at least one bulb to use maximum power.
To solve more question on Power use, visit the link below -
https://brainly.com/question/22103646
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