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?

Related Questions

CBSE CLASS XII Related Questions

1.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

4.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

6.
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.

Similar Physics Concepts

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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

1. The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

4.With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

6.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.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The resistance of a metal wire at \( 20^\circ \text{C} \) is \( 1.05 \, \Omega \) and at \( 100^\circ \text{C} \) is \( 1.38 \, \Omega \). Determine the temperature coefficient of resistivity of this metal.

4.
With the help of a labelled diagram, explain the principle, construction and working of an a.c. generator.

6.
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.