Question

A $500W$ heater is designed to operate at $200V$ potential difference. If it is connected across $160V$ line, the heat it will produce in $20$ minutes is

• A. 384 kJ
• B. 483 kJ
• C. 843 kJ
• D. 348 kJ

Answer 1

Answer: (A)

Case-I Resistance of heater when it operates at $200V$ and power, $P_1=500W$
$\because$ Power, $P=\frac{V^2}{R}$
$\Rightarrow R_1=\frac{V_1^2}{P_1}$
$R_1=\frac{(200{)}^2}{500}=\frac{200\times 200}{500}$
$R_1=80\Omega$
Case - II Now, it connected across $160V$ line the heat it will produce in $20$ minute is
$V_2=160V$
$\therefore P_2=\frac{V_2^2}{R}=\frac{(160{)}^2}{R}$
$=\frac{(160{)}^2}{80}=\frac{160\times 160}{80}=2\times 160(\because R=80\Omega )$
$P_2=320W$
$\therefore$ Now, Heat produce is $20$ minutes
$dQ=P_2\times$ time $=320\times 20\times 60$
$dQ=384kJ$