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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2.
Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

3.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.
This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

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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.Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

3.Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

3.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.
This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?