Question

Two certain appliances A and B are rated as 10kW, 230V and 100W, 230V. Which of the following statement can be regarded correct according to the given data?

• A. appliance A draws less current than B
• B. appliance B has less resistance than A
• C. appliance B generates more heat than A
• D. appliance A has less resistance than B

Answer 1

The correct option is D

appliance A has less resistance than B

Given:
rated power of appliance A = $P_A$ = 10kW = 10000W
rated voltage of appliance A = $V_A$ = 230V
rated power of appliance B = $P_B$ = 100W
rated voltage of appliance B = $V_B$ = 230V
Now to find resistance of appliannce A = $R_A$
$P_A=\frac{{V_A}^2}{R_A}$
so, $R_A=\frac{{V_A}^2}{P_A}$
= $\frac{{\left(230\right)}^2}{10000}$
$R_A$ = 5.29 ohms
similarly $R_B$ = 529 ohms
therefore $R_A$ < $R_B$
Now let current drawn by appliance A = $I_A$
$P_A=\left(I_A{)}^2\right(R_A)$
$I_A^2=\frac{P_A}{R_A}$
$I_A^2=\frac{10000}{5.29}$
$I_A$ = 43.47amperes
let current drawn by appliance B = $I_B$
Similarly $I_B$ = 0.434 amperes
So $I_A$ > $I_B$
Heat generated by appliance A = $H_A$
Heat generated by appliance B = $H_B$
let time t be unity i.e, t = 1s
now $H_A=\left(I_A{)}^2\right(R_A)t$
$H_A$ = ${43.47}^2$ $\times$ 5.29 $\times$ 1
$H_A$ = 9996.2 J
similarly $H_B$ = 99.64 J
therefore $H_A$ > $H_B$
So according to above calculations, the final conclusion are:
$R_A$ < $R_B$
$I_A$ > $I_B$
$H_A$ > $H_B$