Ask question

Ask question

All categories

2 years ago

14

Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house

wiring, what must the ratio of the primary to the secondary turns be in the transformer? (Ignore any changes in voltage due to rectification.)

1 answer:

Vika [28.1K]2 years ago

6 0

Answer:

1/0.0545

Explanation:

The transformer have a ratio of the primary turns to the secondary turns equivalent to the ratio of the voltage it is changing. If we have 6.0 V in the secondary (output), and 110 V in the primary (input) the ratio of the voltage is 6/110 = 0.0545, so the ratio between turns in the transformer is 1/0.0545 (for each turn in the primary, the transformer has 0.0545 turns in secondary)

You might be interested in

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.

loris [4]

<u>Answer</u>

48 Volts

<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
Ns ⇒number of turns in the seconndary coil
Vp ⇒ primary voltage
Vs ⇒secondary voltage

Np/Ns = Vp/Vs

10/4 = 120/Vp

Vp = (120 × 4)/10

= 480/10
= 48 Volts

5 0

2 years ago

Read 2 more answers

Determine the length of a copper wire that has a resistance of 0.172 ? and cross-sectional area of 7.85 × 10-5 m2. The resistivi

KonstantinChe [14]

Answer:

Length of copper wire, l = 785 meters

Explanation:

Given that,

Resistance of the copper wire, R = 0.172 ohms

Area of cross section, $A=7.85\times 10^{-5}\ m^2$

Resistivity of copper, $\rho=1.72\times 10^{-8}\ \Omega-m$

The resistance of a wire is given by :

$R=\rho\dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.172\ \Omega\times 7.85\times 10^{-5}\ m^2}{1.72\times 10^{-8}\ \Omega-m}$

l = 785 meters

So, the length of the copper wire is 785 meters. Hence, this is the required solution.

8 0

2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0

2 years ago

You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m

mamaluj [8]

The correct answer is:
<span>paramagnetism

In fact, paramagnetic materials, when they are placed in a magnetic field, they form an internal magnetic field parallel to the external one and in the same direction. However, unlike ferromagnetic materials, they do not retain their magnetization, so when the external magnetic field is removed, their internal induced magnetic field disappears.</span>

7 0

2 years ago