Question

An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 350 W when the heating is at the minimum. the voltage is 200v what are the current and the resistance in each case.​

Answer 1

Answer:

When heating is at maximum rate (P=840 W)

Current is 3.82A and resistance is 57.61 Ω

When heating is at the minimum rate (P=360 W)

Current is 6.4A and resistance is 134.44 Ω

Explanation:

Given,

Potential difference = V = 220

Maximum Power = 840 W

Minimum Power = 360 W

We know that,

P = VI

∴ $\frac{P}{V}=I$

=>  $I=\frac{P}{V}$

And my Ohm's law,

$V=IR$

∴ R = $\frac{V}{I}$

Finding Current

• For P = 840 W

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

$I=\frac{840}{220}$

$I=\frac{42}{11}$

$I = 3.82 A$

• For P = 360 W

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

$I=\frac{360}{220}$

$I=\frac{18}{11}$

$I = 6.4 A$

Finding Resistance

• For P = 840 W

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

$R=\frac{220}{3.82}$

$R= \frac{220}{\frac{42}{11}}$

$R=\frac{220x11}{42}$

$R=\frac{1210}{21}$

$R=57.61$ Ω

• For P = 360 W

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

$R=\frac{220}{1.64}$

$R=\frac{220}{\frac{18}{11} }$

$R = \frac{110x11}{9}$

$R=134.44$ Ω

When heating is at maximum rate (P=840 W)

Current is 3.82A and resistance is 57.61 Ω

When heating is at the minimum rate (P=360 W)

Current is 6.4A and resistance is 134.44 Ω