Calculating Combined Power of Heaters in Series

The question asks for the combined power when two identical heaters, each rated 220 V, 1000 Watt, are connected in series across a 220-volt line.

First, let's understand the rating of each heater. A rating of 220 V, 1000 Watt means that when connected to a 220 V supply, the heater consumes 1000 W of power. We can use this information to calculate the resistance of each heater.

The relationship between power (P), voltage (V), and resistance (R) is given by the formula:

\(P = \frac{V^2}{R}\)

From this, we can find the resistance (R) of one heater:

\(R = \frac{V^2}{P}\)

Given \(V = 220 \text{ V}\) and \(P = 1000 \text{ W}\):

\(R = \frac{(220 \text{ V})^2}{1000 \text{ W}} = \frac{48400}{1000} = 48.4 \, \Omega\)

So, each identical heater has a resistance of \(48.4 \, \Omega\).

Next, the two heaters are connected in series across a 220 V supply line. When resistors (or heaters in this case) are connected in series, the total resistance is the sum of individual resistances.

\(R_{total} = R_1 + R_2\)

Since both heaters are identical with resistance \(R = 48.4 \, \Omega\), the total resistance is:

\(R_{total} = 48.4 \, \Omega + 48.4 \, \Omega = 96.8 \, \Omega\)

Now we need to find the combined power consumed by the series combination when connected across the 220 V supply line. We can use the same power formula, but this time using the total resistance and the supply voltage:

\(P_{combined} = \frac{V_{supply}^2}{R_{total}}\)

Given \(V_{supply} = 220 \text{ V}\) and \(R_{total} = 96.8 \, \Omega\):

\(P_{combined} = \frac{(220 \text{ V})^2}{96.8 \, \Omega} = \frac{48400}{96.8}\)

Calculating the value:

\(P_{combined} = 500 \text{ Watt}\)

Therefore, the combined power of the two identical heaters when placed in series across a 220 volt line is 500 Watt.

Let's compare this with the original power rating. When connected individually to 220 V, each consumed 1000 W. When connected in series to the same 220 V, the total resistance increases, which leads to a decrease in the total power consumed.

The final answer is 500 Watt.