Question

why voltage is same for both the coils in electric kettle

Answer 1

Heat produced, $H= \large\frac{V^2 t}{IR} $

$\qquad=> R \infty t$

Given $R_1= 6 \Omega$

and $R_2 = 3 \Omega$

$\therefore \large\frac{R_1}{R_2} = \large\frac{6}{3}$

$\qquad= 2\Omega$

$\therefore R_1 = 2R_2$

When the coils are connected in series , the net resistance , $R^1=R_1+R_2=R_2+2R_2=3R_2$

$\therefore \large\frac{R_1}{R_2}$$ =R_2 +2 R_2$

$\qquad= 3 R_2$

$\therefore \large\frac{R1}{R_2} =\frac{t1}{3}$

$t^1= \large\frac{3R^1}{R^2} $

$\qquad= \large\frac{3 \times 3 R_2}{R_2}$

$\qquad= 9 \;m$

Answer : 9 m

Answer 2

This is probably the major one: When the element is at the bottom, one can boil any amount of water that will cover the element. Hot water is less dense cold water, so hot water rises from a hot element to the top of the water, giving off heat all the way, so the heating is uniform. This is called natural convection.

Mark Brainliest...