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3 years ago

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A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac

ross the primary coil has a voltage given by the function Δv = (130 V)sin(ωt). What rms voltage (in V) is measured across the secondary coil?

1 answer:

Sonbull [250]3 years ago

7 0

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

$\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V$

The rms voltage (in V) measured across the secondary coil is calculated as;

$V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V$

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

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2 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

3 years ago

Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f

mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , $h_{f} = 1344.8 kJ/kg$

Specific internal energy of saturated liquid at 300°C, $U_{f1} =$ 1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa, $U_{f2}$ = 340.49 kJ/kg

$(U_{fg} )_{2} = 2142.7 kJ/kg$

Let the dryness fraction at the second chamber = x

$U_{2} = U_{f2} + U_{fg2}$

$U_{2} = 340.49 + x2140.7$ Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, $u_{f2} = 0.00103 m^{3} /kg\\$

$u_{2} = u_{f2} + xu_{fg2}$

$u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\$

$u_{2} = \frac{v_{2} }{m_{2} }$

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C $S_{1} = S_{f} = 3.2548 kJ/kg-k\\$

$S_{2} = S_{f2} + xS_{fg2}$

From the steam table,

$S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k$

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0

3 years ago

Read 2 more answers

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago