Question

Two heated wires of the same dimensions are first connected in series and then it’s parallel to a source of supply. What will be the ratio of heat produced in the two cases?

Answer 1

Here is the step-by-step solution:

The Short Answer

The ratio of heat produced in the series case to the parallel case is 1:4.

---

Detailed Explanation

Let's break down why this is the case.

Step 1: Assumptions and Key Formulas

1. Resistance: Since the two heated wires have the same dimensions and are presumably made of the same material, they will have the same electrical resistance. Let's call the resistance of one wire R.

2. Source of Supply: We assume the "source of supply" is a constant voltage source, like a battery or a wall outlet. Let's call this voltage V. This is a key assumption.

3. Heat and Power: The heat produced ($H$) in a resistor over a period of time ($t$) is given by $H = P \times t$, where $P$ is the power dissipated. The ratio of heat produced in the same amount of time is therefore the same as the ratio of the power dissipated.

4. Power Formula: We can calculate electrical power ($P$) using three main formulas:
   $P = V \times I$
   $P = I^2 \times R$
   * $P = V^2 / R$

Since the voltage ($V$) of the source is constant for both the series and parallel circuits, the most convenient formula to use is P = V² / R_total.

Step 2: Wires Connected in Series

When the two wires are connected in series, the total resistance of the circuit ($R_s$) is the sum of their individual resistances.

• $R_s = R + R = 2R$

Now, we can calculate the power dissipated (which corresponds to the heat produced) in the series circuit ($P_s$):

• $P_s = V^2 / R_s$
• $P_s = V^2 / (2R)$

Step 3: Wires Connected in Parallel

When the two wires are connected in parallel, the reciprocal of the total resistance ($R_p$) is the sum of the reciprocals of their individual resistances.

• $1/R_p = 1/R + 1/R = 2/R$
• Therefore, $R_p = R/2$

Now, we can calculate the power dissipated in the parallel circuit ($P_p$):

• $P_p = V^2 / R_p$
• $P_p = V^2 / (R/2)$
• $P_p = 2V^2 / R$

Step 4: Finding the Ratio

The question asks for the ratio of heat produced in the two cases (Series : Parallel). This is the ratio $P_s : P_p$.

• Ratio = $P_s / P_p$
• Ratio = $(V^2 / 2R) / (2V^2 / R)$

To simplify this fraction, we can invert the denominator and multiply:

• Ratio = $(V^2 / 2R) \times (R / 2V^2)$

Now, we cancel out the common terms ($V^2$ and $R$):

• Ratio = $1 / (2 \times 2)$
• Ratio = 1/4

Conclusion

The ratio of the heat produced when the wires are in series to when they are in parallel is 1:4.

Intuitive Reason: In the parallel configuration, the total resistance of the circuit is much lower ($R/2$) compared to the series configuration ($2R$). According to Ohm's Law ($I = V/R$), a lower resistance allows a much higher total current to flow from the constant voltage source. Since power ($P=V \times I$) is directly proportional to this current, the parallel circuit draws significantly more power and therefore produces significantly more heat.