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poizon [28]
1 year ago
10

A 3. 5-a current is maintained in a simple circuit with a total resistance of 1500 ω. What net charge passes through any point i

n the circuit during a thirty second interval?.
Physics
1 answer:
melamori03 [73]1 year ago
7 0

Electric Current:

Electric current is the flow of charge through a given circuit per unit time. Electric current is one of the components needed to calculate the electric power that a device needs to operate and do work. Electric current is measured in amperes (A), which is equal to:

            1A = 1 C/ s

Recall that the coulomb (C) is the unit for charge while the second (s) is the unit for time

Given: I = 3.5

A is the current

Δt =30 s is the time interval

                     A =ΔQ/ΔT

Net charge = 100C

Electricity is produced when an electric current runs through a circuit.

How does electric current work?

A current of electricity is a steady flow of electrons. When electrons move from one place to another, round a circuit, they carry electrical energy from place to place like marching ants carrying leaves. Instead of carrying leaves, electrons carry a tiny amount of electric charge.

Learn more about Electric current :

brainly.com/question/27003377

#SPJ4

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so here if distance is decreased the force will increase'

now if force is tripled

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In this circuit (see picture), which resistor will draw the least power?
Basile [38]
A few different ways to do this: 

Way #1: 
The current in the series loop is  (12 V) / (total resistance) . 
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is  (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
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And by the way, it's not "drawing" the most power.  It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

                 (voltage across the resistance)²  /  (resistance)  .

In a series loop, the voltage across each resistor is

          [ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

         (individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .                                      

Way #3:  (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===>  When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
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What are the two forces involved in an interaction called
natka813 [3]

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Explanation:

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Answer:

B

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