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3 years ago

13

The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 4.5 wb from 13.0 wb in a time of 0.072 s. th

e average induced current in the coil is 190
a. what is the resistance of the wire?

1 answer:

zavuch27 [327]3 years ago

8 0

We could answer the question through the help of Faraday's Law of Induction:

Volts induced = - N•dΦ/dt

where N is the number of turns, and

Φ is the magnitude of the magnetic field.

V = IR, or R = V / I

V = -18 * [13Wb – 4.5Wb] / 0.072s

V = -2125 volts the sign just specifies direction

R = -2125V / 190 A

R = -11.1842105 Ω

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A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

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<h2>Answer:</h2>

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<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

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<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

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R₁ = resistance in the first bulb

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Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

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Where;

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To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

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Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

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Following the same approach, to get R₂, equation (ii) can be written as;

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Where;

V = 12.0V

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Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

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12.0 = 11.7 + 0.65r

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0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

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