Question

Current flows in a conductor has a resistance 2 ohm is increased by 4ohm thend the rate of flow of current is
2 times
4 times
O.5 times
0.25 times

Answer 1

According to the Ohm's Law ,

Current flowing through the conductor is directly proportional to the potential difference applied across the ends provided that physical conditions are constant .

In the given problem , resistance is increased from 2 Ω to 4 Ω .

There will be a change in the current for sure , but if both current and resistance changes , then the constant should be potential difference applied across both the ends .

According to Ohm's Law , we can write :

V ∝ I

⇒ V = I R such that resistance is constant .

Now that V is constant :

I = V / R

The initial current was I and the initial resistance was 2 Ω ,

I = V / 2

Now the resistance was increased to 4 Ω .

This means that now the resistance becomes 4 Ω  .

Thus the current becomes I = V / 4

The original current was V / 2 and not it becomes V / 4 .

Ratio = ( V/2 ) / ( V/4 )

⇒ ( 1/2 ) / ( 1/4)

⇒ ( 0.5 ) / ( 0.25 )

⇒ 2

Hence the original current was twice the final current .

Let original current be i and the final be I .

i = 2 I

⇒ I = i / 2

⇒ I = 1/2 × i

⇒ I = 0.5 i

The rate of flow of current becomes 0.5 times the original flow .

Answer 2

Current flows in a conductor has a resistance 2 ohm is increased by 4ohm thend the rate of flow of current is.

Since resistance doubled here from 2 ohm to 4 ohm.

I = V/R

I - current,

V - voltage,

R - resistance.

Current is inversely proportional to resistance.

If resistance is doubled, then current will be halved.

In simple way,

As voltage is constant.

V = IR - - eq1)

V = I'*2R (second case where resistance doubled) - - eq2)

By comparing one and two,

=) I * R = I' * 2R

=) I/2 = I'

0.5 times of initial current.

Hence option C is correct.