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dusya [7]
2 years ago
11

Please solve the Problem.

Physics
1 answer:
STatiana [176]2 years ago
5 0

(a) For series circuit, current in 14 ohms = current in 72 ohms = 0.698 A

(b) Power loss in each series resistor, 14 ohms = 6.82 W and 72 ohms = 35.1 W.

(c) For parallel circuit, current in 14 ohms = 4.29 A and current in 72 ohms = 0.83 A

(d) Power loss in each parallel resistor, 14 ohms = 257.4 W and 72 ohms = 49.8 W.

<h3>Current in each series resistors</h3>

The total resistance = R1 + R2

                              R = 14 + 72 = 86 ohms

Current = V/R = 60/86 = 0.698 A

Since the resistors are in series, current in 14 ohms = current in 72 ohms = 0.698 A

<h3>Power loss in each series resistor</h3>

P = I²R

P(14 ohms) = (0.698)² x 14 = 6.82 W

P(72 ohms) = (0.698)² x 72 = 35.1 W

<h3>Current in each resistor for parallel arrangement</h3>

Total resistance, 1/R = 1/R1 + 1/R2

1/R = 1/14 + 1/72

1/R = 0.0853

R = 1/0.0853

R = 11.72

Total current in the circuit = V/R = 60/11.72 = 5.12 A

Current in 14 ohms = 60/14 = 4.29 A

Current in 72 ohms = 60/72 = 0.83 A

<h3>Power loss in each parallel resistor</h3>

P = IV

P(14 ohms) = (4.29) x 60 = 257.4 W

P(72 ohms) = 0.83 x 60 = 49.8 W

Learn more about current here: brainly.com/question/24858512

#SPJ1

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Therefore,

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Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

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