Question

An electric iron consumes energy at the rate of 880W when heating is at the maximum rate and 440W when the heating is at minimum rate .the applied voltage is 220V calculate current and resistance

Answer 1

Answer:

Given :

Power at maximum rate =P=880W

voltage =V=220volts.

We know that P=V²/R

R=V²/P

⇒220x220/880

⇒48400/880

⇒55ohms

Maximum current I is

P=IV

I=P/V

=880/220

=4A

ii)power at minimum rate is 440W

voltage =V=220volts.

We know that P=V²/R

R=V²/P

⇒220x220/440

⇒48400/440

≈19ohms

Current at minimum rate is ,

P=IV

I=P/V

⇒440/220

⇒2A

Answer 2

Answer:

Current is R=55 $\Omega$ and resistance is R=110 $\Omega$.

Explanation:

The electric iron operates under the premise that current has a heating impact. When electricity passes through the metal in the electric iron, the metal is heated. All of the metal's electrons are energized and move as a result of the electric current passing through it.

The main work of an electric iron is to heat up. So the main principle behind an electric iron is heating of material through current. The material used in it is a good conductor of heat which conducts heat evenly on its surface.

An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when the heating is at the minimum. The voltage is 220 V.

Given :

Power at maximum rate =P=880W

voltage =V=220volts.

We know that P=V²/R

Where V is voltage, I is current , R is resistance.

Step 1:

First case,

Power $=880 \mathrm{~W} \Rightarrow 880=220 \times I \Rightarrow I=4 \mathrm{~A}$

$\Rightarrow 880=\frac{220^2}{\mathrm{R}} \Rightarrow \mathrm{R}=55 \Omega$

Step 2: power at minimum rate is 440W

voltage =V=220volts.

We know that P=V²/R

R=V²/P

$\begin{aligned}& \text { Power }=440 \mathrm{~W}=220 \times \mathrm{I} \Rightarrow \mathrm{I}=2 \mathrm{~A} \\& \Rightarrow 440=\frac{220^2}{\mathrm{R}} \Rightarrow \mathrm{R}=110 \Omega\end{aligned}$

Thus, Current is R=55 $\Omega$ and resistance is R=110 $\Omega$.

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