Question

An electric iron consumes energy at a rate of 840 W when heating
is at the maximum rate and 360 W when the heating is at the
minimum. The voltage is 220 V. What are the current and the
resistance in each case?​

Answer 1

Answer:

The answer will be :

In Case I : I = 3.8 A approx.

                R = 57.89 ≈ 58 ohms;

In Case I : I = 1.6 A approx.

                R = 137.5 ohms;

Explanation:

Since it is the  case of per second, so we will take time as 1 second (if necessary then, otherwise no) and other as their respective values.

Now, at maximum rate, power consumed = 840W.

       at minimum rate the power consumed will be = 360 W.

          The potential difference given is = 220 V.

Now, we know that Power or P = VI ;                           (i)

 Case I ( maximum rate) : P = 840 W;

                                          V = 220 V;

              So, after applying the formula in equation (i), we get;

                            840 = 220 (I);

                             I = 840 / 220;

                             I = 42 / 11;

                             I = 3.8 A approx.

             Now, current or I = 33.8 A;

                        V = 220 V;

           So, according to ohms law (which states V = IR), R will be;

                       R = 220 / 3.8 ;

                          = 57.89 ≈ 58 ohms;

Now, Case II ( minimum rate) : P = 360 W;

                                                  V = 220 W;

             So, after applying formula in equation (i), we get;

                              360 = 220 (I);

                                I = 360 / 220;

                                I = 18 / 11;

                                  =  1.6 A approx.

           So, according to ohms law (which states V = IR), R will be;

                       R = 220 / 1.6 ;

                          = 137.5 ohms;

That's all.

Answer 2

Answer:

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