Question

please answer this question ​

Answer 1

Answer:

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Answer 2

$\huge\fbox\pink {A}\fbox\blue{n}\fbox\purple {s}\fbox\green {w}\fbox\red {e}\fbox\orange {r}\fbox\gray {:-}$

The amount of the current flowing through the electric component is decreased by the half.

$\huge{\red {\underline {\overline {\mathfrak {Explanation:-}}}}}$

$V' = \frac {V}{2}$

$I' = \frac {V'}{R}$

R = R

$\huge {According\:to\:Ohm's\:Law}$

V= IR

I = $\frac {V}{R}$

$\huge {So\:,}$

$\frac {V}{R} = \frac {\frac {V}{2}}{R}$

$\frac {V}{R} = \frac {1}{2}\times\frac {V}{R}$

$\frac {V}{R}\times\frac {R}{V} = \frac {1}{2}\times1$

$= \frac{1}{2}$