Question

Two heaters A and B having power rating 3 kW, 220 V and 6 kW, 220 V respectively when connected in series with 220 V supply, boils the water in 9 second. When these two heaters are connected in parallel with 220 V supply, will boil the water in

• A. 2 s
• B. 3 s
• C. 4 s
• D. 5 s

Answer 1

Answer: (A)

2 s

Answer 2

Solution:

1. Calculate Resistances:

   $$R_A = \frac{V^2}{P_A} = \frac{220^2}{3000} \approx 16.13\,\Omega, \quad R_B = \frac{V^2}{P_B} = \frac{220^2}{6000} \approx 8.07\,\Omega$$

2. Series Connection:

   $$R_{\text{series}} = R_A + R_B \approx 16.13 + 8.07 = 24.20\,\Omega$$ $$P_{\text{series}} = \frac{220^2}{24.20} \approx 2000\,\text{W}$$

   Energy required to boil water:

   $$Q = P_{\text{series}} \times 9 = 2000 \times 9 = 18000\,\text{J}$$

3. Parallel Connection:

   When connected in parallel, each heater gets full 220 V:

   $$P_A = 3000\,\text{W},\quad P_B = 6000\,\text{W} \quad \Rightarrow \quad P_{\text{parallel}} = 3000 + 6000 = 9000\,\text{W}$$

   Time required:

   $$t = \frac{Q}{P_{\text{parallel}}} = \frac{18000}{9000} = 2\,\text{s}$$

Summary:

• Calculated resistances using $R = \frac{V^2}{P}$.
• In series, power is $\frac{220^2}{R_A + R_B}$ resulting in 2000 W, giving an energy requirement of 18000 J over 9 s.
• In parallel, total power is 9000 W, so time required is $18000/9000 = 2$ s.