Question

An electric iron consumes energy at a rate of 840w when heating is at the max rate and 360w when heating is at minimum.the voltage is 220v,what are the current and resistance in each case

Answer 1

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Answer 2

Given :-

• Power of Iron when hearting rate is maximum = 840 watt

• Power of Iron when heating rate is minimum = 360 watt

• Voltage (V) = 220 V

To find :-

We have To calculate the Current and resistance when power rate is Maximum and minimum

So , Formula used !

• To find current

$\underline{\boxed{\sf { Power (P) = Voltage (V) \times Current (I) }}}$

• To find Resistance !

$\underline{\boxed{\sf { Voltage (V) = Current (I) \times Resistance (R) }}}$

$\underline{\small{\sf{\dag \ \ Calculate \ Current \ and \ Resistance\ when \ power \ is Maximum}}}$

$\sf\ \ P_m= VI_1\\ \\ :\implies\sf 840 = 220\times I_1\\ \\ :\implies\sf \cancel{\dfrac{840}{220}}= I_1\\ \\:\implies\sf 3.81= I_1$

$\boxed{\sf{\dag \ \ Current (I_1)= 3.81 \ A \ when (Power\ is \ maximum}}$

• Calculate Resistance

$:\implies\sf V= I_1R_1 \\ \\ :\implies\sf 220= 3.81 \times R\\ \\:\implies\sf \cancel{\dfrac{220}{3.81}}= R\\ \\ :\implies\sf 57.74 = R_1$

$\boxed{\sf{\dag \ \ Resistance (R_1) = 57.74 \Omega \ when (Power\ is \ maximum}}$

$\underline{\small{\sf{\dag \ \ Calculate \ Current \ and \ Resistance\ when \ power \ is Minimum}}}$

$\sf\ \ P_n= VI_2\\ \\ :\implies\sf 360= 220\times I_2\\ \\ :\implies\sf \cancel{\dfrac{360}{220}}= I_2\\ \\:\implies\sf 1.63= I_1$

$\boxed{\sf{\dag \ \ Current (I_1)= 1.63 \ A \ when (Power\ is \ minimum}}$

• Now Resistance

$:\implies\sf V= I_2R_2 \\ \\ :\implies\sf 220= 1.63 \times R\\ \\:\implies\sf \cancel{\dfrac{220}{1.63}}= R\\ \\ :\implies\sf 134.9 = R_1$

$\boxed{\sf{\dag \ \ Resistance (R_1) = 134.9 \Omega \ when (Power\ is \ minimum}}$