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3 years ago

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2) What must be connected for an electrical current to flow in a simple current?

1 answer:

hjlf3 years ago

6 0

2) What must be connected for an electrical current to flow in a simple current?

<em>d.moving magnets,conductors, load to do work </em>

here moving magnets will work as generator and conductors is used as connectors.

5)What is the purpose of a load in a circuit?

<em>c.provide resistance to keep the voltage strong </em>

Load will provide the resistance to the circuit and help to build the voltage across it

6) What cause an electric current from a generator

<em>d.Magnets and conducting wires move near each other </em>

here when magnets move near wires then it will induce the electric current and behaves like generators

7) What do all generators have in common

<em>b.they transform mechanical into electrical energy </em>

They have magnets and rotating coils which will convert magnetic energy to electrical energy

8) How can we measure electrical current

<em>b.the pressure force is voltage;the # of electrons moving is amperes </em>

this voltage and number of electrons will help to calculate current

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Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr

slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

$E=\frac{1}{2} *L*I^{2}$

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

$E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}$

Then:

$E'=4*E$

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

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