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2 years ago

The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.

Physics

1 answer:

zzz [600]2 years ago

7 0

In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

Simple SPST

(ON)-OFF, Push-to-close, SPST Momentary

ON-(OFF), Push-to-Open, SPST Momentary

Inches Switch SPST

Learn more about terminal here:

brainly.com/question/14236970

#SPJ4

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A long wire carrying a 5.0 A current perpendicular to the (xy)-plane intersects the x-axis at x = -2.00 cm. A second parallel wi

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Answer:

a. 05cm from x axis

b. 8cm from x axis

Explanation:

If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below

a. Given that i1= 5A and i2=3A

Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.

Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(4-x)

i1/x=i2/(4-x)

5/x=3/(4-x)

20-5x=3x

8x=20

8x=2.5cm

since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.

The null point is 0.5cm from the origin x axis

b. If both current are flowing in opposite direction, the null point lies outside of the current.

Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.

Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point

B1=B2

μi1/2πx=μi2/2π(x-4)

i1/x=i2/(x-4)

5/x=3/(x-4)

5x-20=3x

2x=20

x=10cm.

This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.

The null point is 8cm from the x axis.

Check the attachment to see the diagram of the current and the null points

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