Question

Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) Half (b) Double (c) One-fourth (d) Four times

Answer 1

Answer:

As from Ohm's Law,

V = IR........1

Now, here V is constant.

Let , new resistance R' = 2R and assume that current flow be I'

So, from 1,

IR = I'.R',

-> IR = I'.2R,

Therefore, I' = I / 2

So the current is halved.

Explanation:

Pls mark me as brainlist.

Answer 2

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

Keeping the potential difference constant, the resistance of a circuit is doubled.

$\checkmark$ To Find:

The current will become

• Half
• Double
• One fourth
• Four times

$\checkmark$ Solution:

We know that,

$\star$ Ohm's law:

Current is directly proportional to Voltage and Inversely proportional to Resistance.

Mathematically,

$\blue{\implies \rm I \propto V}$

$\green{\rm \implies I \propto \dfrac{1}{R}}$

Therefore,

$\red{\boxed{\rm V=IR}}$

Hence, Current ($\rm I$)

$\orange{\implies \rm I=\dfrac{V}{R}}$

Here,

V = Voltage

$\rm I$ = Current

R = Resistance

According to Question,

Given that,Keeping the potential difference constant, the resistance of a circuit is doubled.

Therefore,

$\blue{\implies \rm R'=2R}$

Hence,

The New Current ($\rm I'$)

$\green{\implies \rm V=I'R'}$

$\red{\implies \rm V=I'(2R)}$

$\orange{\rm \implies I'=\dfrac{1}{2}\left(\dfrac{V}{R}\right)}$

We know that,

$\orange{\implies \rm I=\dfrac{V}{R}}$

Therefore,

$\orange{\rm \implies I'=\dfrac{1}{2}(I)}$

Hence,

The Current will become Half.

Therefore,

$\checkmark$ Option (a) is correct