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3 years ago

List the basic components of a simple electric circuit.Explain why each of the components is necessary

Physics

1 answer:

Alex_Xolod [135]3 years ago

6 0

You should check out this website:

https://www.hunker.com/12003706/the-four-and-more-basic-parts-of-an-electrical-circuit

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Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger

Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

4 0

3 years ago

To identify whether the forces are balanced or not, you need to whet

jok3333 [9.3K]

Answer:

what are you asking?

Explanation:

7 0

3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

3 years ago

A model train traveling at a constant speed around a circular track has a constant velocity

Mrac [35]

The answer should be yes

7 0

3 years ago

Read 2 more answers

2.Cars were previously manufactured to be as sturdy as possible, whereas today's cars

Marina CMI [18]

Answer:

Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.

4 0

3 years ago