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2 years ago

Explain about ohm's law.

Physics

2 answers:

belka [17]2 years ago

4 0

Answer:

Statement:

The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.

Explanation:

Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).

V∝I or I∝V⇒V=IR.

Anastaziya [24]2 years ago

4 0

Explanation:

Ohm's law states that the current through a conductor is proportional to the voltage across the conductor. This is true for many materials (including metals) provided the temperature (and other physical factors) remain constant. The constant of proportionality, R,R is the resistance and the unit is the ohm, with symbol \Omega,Ω. The relationship can be written as:

V, equals, I, R.

V=IR

where V,V is the voltage across the conductor and I,I is the current flowing through it. If a component is ohmic (it obeys Ohm's Law), then its resistance must be independent of current and voltage.

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Neko [114]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

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here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

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also the acceleration of the stone is due to gravity which is given as

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now we have

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so from above equation

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ivanzaharov [21]

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2 years ago

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dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

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T is time of motion
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1 year ago

Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu

Natasha2012 [34]

1) Current in each bulb: 0.1 A

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$R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega$

And so, the current through the circuit is (using Ohm's law):

$I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A$

And since the two bulbs are connected in series, the current through each bulb is the same.

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The power dissipated by each bulb is given by the formula:

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For the first bulb:

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For the second bulb:

$P_1 = (0.1 A)^2 (800 \Omega)=8 W$

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2 years ago

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3 years ago

Explain about ohm's law.​

Explain about ohm's law.