Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

According to Ohm’s law

V = IR

where

V is the potential difference
I is Current
R is Resistance

The equation can also be written as –

I = V/R ---- (1)

It is given that the potential difference across two ends of an electric component decreases to half

So, the new potential difference be Vʹ= V/2---- (2)

Resistance remains constant, therefore the new current that is drawn through the electrical component is

Iʹ = Vʹ/R ---- (3)

Substituting the value of V’ from equation (2) in equation (3), we get

I’ = (V/2)/R

I’ = (1/2) (V/R)

Since V/R = I

I’ = (1/2) I

I’ = I/2

Therefore, the amount of current that flows through the electrical component is reduced by half.

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Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

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A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

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Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Related Questions

CBSE CLASS XII Related Questions

2. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

4.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

5.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

6. Write any two features of nuclear forces.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

3.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

5.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

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