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Vlada [557]
3 years ago
14

A transformer consisting of two coils wrapped around an iron core is connected to a generator and a resistor (Resistor is connec

ted to the secondary coil). There are 11 turns in the primary coil and 18 turns in the secondary coil. The peak voltage across the resistor is 67 V. What is the peak emf of the generator?
Physics
1 answer:
V125BC [204]3 years ago
4 0

Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

Peak voltage = 67 V

We nee to calculate the peak emf

Using relation of number of turns and emf

\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}

E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

E₂ = emf across secondary coil

Put the value into the formula

E_{1}=\dfrac{11}{18}\times67

E_{1}=40.94\ V

Hence, The peak emf of the generator is 40.94 V.

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Read 2 more answers
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a
seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

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½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

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