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3 years ago

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Two resistors of 2 ohms each are connected in parallel, another resistor of 1 ohm is connected in series with the parallel combi

nation, and the resistive combination is connected to a 2 volt source. How much power is dissipated in either one of the parallel resistors?

1 answer:

insens350 [35]3 years ago

3 0

Answer:

The power dissipated in either one of the parallel resistors is 2 V

Explanation:

Given;

two parallel resistors, R₁ and R₂ = 2 ohms

The total resistance of the Two resistors of 2 ohms connected in parallel is;

$R_T = \frac{R_1R_2}{R_1+R_2} = \frac{2*2}{2+2} = \frac{4}{4} = 1 \ ohm$

when connected to another resistor of 1 ohm in series, the total resistance becomes;

Rt = R₁ + R₂

Rt = 1 + 1 = 2 ohms

Current in the circuit, I = voltage / total resistance

= 2 /2 = 1 A

the overall circuit has been resolved to series connection, and current flow in series circuit is constant.

Power = I²R

Thus, power dissipated in either one of the parallel 2 ohms resistors is;

Power = I²R = (1)² x 2 = 2 V

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Read 2 more answers

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=> $T = 2* 34 + 2 + 2$

=> [tex]T =72 \ years /tex]

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