Question

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Answer 1

Question :-

If resistance of a circuit is doubled and P.D is halved, then find the percent increase or decrease on current.

Given :-

Resistance is doubled,

= 2R

P.d is halved,

= 1/2V

Now,

Ohm's law ↦ V = IR

or, I = V/R

New current,

$\sf I=\frac{\frac{1}{2}V }{2R} \\ I = \frac{\frac{v}{2} }{2R} \\ I =\frac{V}{2 \: \times \: 2R} \\ I = \frac{V}{4R} \\ I= \frac{1}{4} \: \times \: \frac{V}{R} \\ I=\frac{1}{4} \: \times \: i \\ I=\frac{1}{4} i$

Now,

percentage (increase)

= (increase in current)/original current x 100

$\sf = \frac{\frac{1}{4} \: \times \: \frac{V}{R} }{\frac{V}{R} } \: \times \: 100\\= \: \frac{1}{4} \: \times \: 100\\=\:25$

Therefore, there is an increase of current by 25%.

Answer 2

Answer:

Question :-

If resistance of a circuit is doubled and P.D is halved, then find the percent increase or decrease on current.

Given :-

Resistance is doubled,

= 2R

P.d is halved,

= 1/2V

Now,

Ohm's law ↦ V = IR

or, I = V/R

New current,

\begin{lgathered}\sf I=\frac{\frac{1}{2}V }{2R} \\ I = \frac{\frac{v}{2} }{2R} \\ I =\frac{V}{2 \: \times \: 2R} \\ I = \frac{V}{4R} \\ I= \frac{1}{4} \: \times \: \frac{V}{R} \\ I=\frac{1}{4} \: \times \: i \\ I=\frac{1}{4} i\end{lgathered}

I=

2R

2

1

V

I=

2R

2

v

I=

2×2R

V

I=

4R

V

I=

4

1

×

R

V

I=

4

1

×i

I=

4

1

i

Now,

percentage (increase)

= (increase in current)/original current x 100

\begin{lgathered}\sf = \frac{\frac{1}{4} \: \times \: \frac{V}{R} }{\frac{V}{R} } \: \times \: 100\\= \: \frac{1}{4} \: \times \: 100\\=\:25\end{lgathered}

=

R

V

4

1

×

R

V

×100

=

4

1

×100

=25

Therefore, there is an increase of current by 25%.