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Inessa05 [86]
3 years ago
8

A step-down transformer has more loops on which coil?

Physics
2 answers:
Nesterboy [21]3 years ago
6 0
A step-down transformer has more loops in :  A. Primary coil

Primary coil refers to the coil to which alternating voltage is supplied. It's usually connected to the AC supply

hope this helps
MrRissso [65]3 years ago
3 0

Answer:

A). Primary coil

Explanation:

As we know that in an ideal transformer the flux linked with primary coil and secondary coil for each turn must be same

so we can say that

\phi _p = N_p \phi_0

\phi_s = N_s\phi_0

so as per faraday's law we can say that two EMF is given as

E_s = N_s\frac{d\phi_0}{dt}

E_p = N_pfrac{d\phi_0}{dt}

now we have

\frac{E_s}{E_p} = \frac{N_s}{N_p}

so we can say that

for step down transformer the output voltage is less than the input transformer

E_s < E_p

so we will have

N_s < N_p

so turns will be more in

A) Primary coil

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Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

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Answer:

Explanation:

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iren [92.7K]

Answer:

angle minimum   θ = 41.3º

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Answer:

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