1. Let the same amount of heat energy to be produced be $H$.

2. When a coil is connected to the same voltage source, the heat produced in time $t$ is $H = Pt$ where $P$ is the power consumed.

3. For the first coil: $H = P_1 \times 20$ so, $P_1 = \frac{H}{20}$

4. For the second coil: $H = P_2 \times 60$ so, $P_2 = \frac{H}{60}$

5. When the two coils are connected in parallel across the same source, total power is the sum: $P_{\text{parallel}} = P_1 + P_2 = \frac{H}{20} + \frac{H}{60}$

6. Take LCM: $P_{\text{parallel}} = H\left(\frac{3+1}{60}\right) = \frac{4H}{60} = \frac{H}{15}$

7. If the combination produces the same heat $H$ in time $t$, then $H = P_{\text{parallel}} \cdot t = \frac{H}{15} \cdot t$

8. Cancelling $H$: $1 = \frac{t}{15}$ $t = 15 \text{ min}$

9. Therefore, the required time is $\boxed{15}$

Comparison with stored answer:

• Stored correct answer: $15$
• Derived answer: $15$
• They match.