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2 years ago

11

A circuit supplied with 110V carries 5 Amps current. Calculate the Resistance value of the circuit?

1 answer:

IgorC [24]2 years ago

6 0

Answer:

The value is $R = 22 \ \Omega$

Explanation:

From the question we are told that

The voltage is $V = 110 \ V$

The current is $I = 5 \ A$

Generally the resistance value is mathematically represented as

$R = \frac{V}{I}$

=> $R = \frac{110}{5}$

=> $R = 22 \ \Omega$

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A sail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 N. how much w

erma4kov [3.2K]

<h3><u>Answer;</u></h3>

a) 5.00 x 10^8 J

<h3><u>Explanation;</u></h3>

The work done to move the sailboat is calculated through the equation;

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Substituting the known values from the given above,

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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

mart [117]

Answer:

The value is $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

The distance of separation is $d = 0.6 \ m$

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Generally the magnitude of the field exerted between the current carrying wire is

$B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

$B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is

$d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

$B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is

$d_2 = 0.3 \ m$

This means that $d_1 = d_2 = a = 0.3$

So

$B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=> $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=> $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

=> $B = 3.33 *10^{-6} \ T$

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Answer:

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