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Jet001 [13]
2 years ago
12

A circuit is supplied with a constant voltage. As the resistance of the circuit decreases, the power given to the circuit

Physics
1 answer:
Yakvenalex [24]2 years ago
7 0

Here, we are required to determine the reaction of the power given to a circuit supplied with a constant voltage as the distance of the circuit decreases.

  • In a circuit supposed with a <em>constant voltage</em>, as the resistance of the circuit decreases, the power given to the circuit Increases as evident in the formular: P = V²/R

The power given to a circuit is given by the formula;

Power, P = V × I

where, V = voltage supplied

and I = Current through the circuit.

However, from Ohm's law;

V = I × R

In essence, I = V/R

By substituting, I = V/R into the equation P = V × I

Therefore, we have;

P = V²/R.

Ultimately, if the circuit is supplied with a constant voltage, As the resistance of the circuit decreases, the power given to the circuit Increases.

Read more:

brainly.com/question/16488641

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A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme
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The component of the force in negative z-direction is -0.144 N.

The given parameters;

  • <em>current in the wire, I = 2.7 A</em>
  • <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
  • <em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

F = ILBsin(\theta)

where;

  • <em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0

The magnitude of the wire length is calculated as follows;

|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m

The component of the force in negative z-direction is calculated as;

F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times  sin(53.3)\\\\F_z = -0.144 \ N

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

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