Question

Two heaters A and B are in parallel across the supply voltage. Heater A produces $500kJ$ in $20$ minutes and B produces $1000kJ$ in $10$ minutes. The resistance of A is $100\Omega$. If the same heaters are connected in series across the same voltage, the heat produced in $5$ minutes will be:

• A. $200kJ$
• B. $100kJ$
• C. $50kJ$
• D. $10kJ$

Answer 1

The correct option is B $100kJ$

$\frac{V^2}{R_1}t=500$ and $\frac{V^2}{R_2}\left(\frac{t}{2}\right)=1000$

$\Rightarrow \frac{2R_2}{R_1}=\frac{1}{2}\Rightarrow R_2=25\Omega$

When the heaters are connected in series, heat produced

$=\frac{V^2}{125}\times 5\times 60$

$=\frac{500\times {10}^3}{25\times 2\times 6}\times 60={10}^{5}J$

$=100kJ$