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murzikaleks [220]
3 years ago
7

A generator consists of a rectangular loop with turns of wire spinning at in a uniform magnetic field. The generator output is c

onnected to a series RC circuit consisting of a and a capacitor. What is the average power delivered to the circuit?
Physics
1 answer:
iren [92.7K]3 years ago
5 0
<h3><u>Full question:</u></h3>

A generator consists of a 18-cm by 12-cm rectangular loop with 500 turns of wire spinning at 60 Hz in a 25 mT uniform magnetic field. The generator output is connected to a series RC circuit consisting of a 150 Ω and a 35 μF capacitor.What is the average power delivered to the circuit?

<h3><u>Answer:</u></h3>

The average power delivered to the circiut is 27.5 W

<h3><u>Solution:</u></h3>

Induced emf amplitude = N A B w  

N- Number of turns of the coil

B- Magnetic field

=500 \times(0.18 \times 0.12) \times 0.025 \times(2 \pi 60)

Induced emf amplitude = 101.8 Volt

\begin{aligned}&X_{c}=\frac{1}{(2 p i f C)}=\frac{1}{2 \times \pi \times 60 \times 35 \times 10^{-6}}\\&X_{c}=75.8\ \mathrm{ohm}\end{aligned}

R = 150 ohm

\begin{aligned}&z=\sqrt{\left[R^{2}+X_{c}^{2}\right]}=168 \text { ohm }\\&I_{\text {peak}}=\frac{101.8}{168}=0.606\ \mathrm{A}\\&P_{a v g}=\frac{l^{2} R}{2}=27.5\ \mathrm{W}\end{aligned}

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Explanation:

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E₁: Mechanical energy at the end (bottom of the incline)

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If v₀ = 0  ⇒  K₀

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we get

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we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

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finally, we can say that

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