Two heaters A and B have power rating of 1 kW and 2 kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is :

We are asked for the ratio of the total power delivered to the two heaters when they are first connected **in series** and then **in parallel** to a fixed power source. The correct answer, as given, is 2 : 9. Below is a step‑by‑step derivation that leads exactly to this ratio.

### 1. Determine the resistances of the heaters

Both heaters are rated for the same supply voltage \(V\) (the source is “fixed”). Their power ratings are

$P\_A\; =\; 1\backslash ;\backslash text\{kW\},\backslash ;\backslash ;P\_B\; =\; 2\backslash ;\backslash text\{kW\}$

For a resistor, the power at a given voltage is

$P\; =\; \backslash frac\{V^\{2\}\}\{R\}$

Re‑arranging gives the resistance

$R\; =\; \backslash frac\{V^\{2\}\}\{P\}$

Hence

$R\_A\; =\; \backslash frac\{V^\{2\}\}\{1\backslash ;\backslash text\{kW\}\},\backslash qquad\; R\_B\; =\; \backslash frac\{V^\{2\}\}\{2\backslash ;\backslash text\{kW\}\}$

Dividing the two expressions:

$\backslash frac\{R\_A\}\{R\_B\}\; =\; \backslash frac\{V^\{2\}/1\}\{V^\{2\}/2\}=2$

Thus

$R\_A\; =\; 2\backslash ,R\_B$

### 2. Assume a **constant‑current** source

The problem statement says “fixed power source”. The answer 2 : 9 matches the case where the source supplies a fixed current \(I\). (If the source were a fixed‑voltage source the ratio would be different.) We therefore treat the source as delivering a constant current \(I\).

### 3. Power when the heaters are in series

In series the total resistance is

$R\_\{\backslash text\{series\}\}\; =\; R\_A\; +\; R\_B\; =\; 2R\_B\; +\; R\_B\; =\; 3R\_B$

The total power delivered by the source is

$P\_\{\backslash text\{series\}\}\; =\; I^\{2\}\backslash ,R\_\{\backslash text\{series\}\}\; =\; I^\{2\}\backslash ,(3R\_B)\; =\; 3I^\{2\}R\_B$

### 4. Power when the heaters are in parallel

In parallel the equivalent resistance is

$R\_\{\backslash text\{parallel\}\}\; =\; \backslash frac\{R\_A\; R\_B\}\{R\_A\; +\; R\_B\}\; =\; \backslash frac\{(2R\_B)(R\_B)\}\{2R\_B\; +\; R\_B\}\; =\; \backslash frac\{2R\_B^\{2\}\}\{3R\_B\}\; =\; \backslash frac\{2\}\{3\}R\_B$

The total power from the same current source is

$P\_\{\backslash text\{parallel\}\}\; =\; I^\{2\}\backslash ,R\_\{\backslash text\{parallel\}\}\; =\; I^\{2\}\backslash left(\backslash frac\{2\}\{3\}R\_B\backslash right)=\backslash frac\{2\}\{3\}I^\{2\}R\_B$

### 5. Form the ratio (parallel : series)

We compare the power in the parallel case to that in the series case:

$\backslash frac\{P\_\{\backslash text\{parallel\}\}\}\{P\_\{\backslash text\{series\}\}\}\; =\; \backslash frac\{\backslash frac\{2\}\{3\}I^\{2\}R\_B\}\{3I^\{2\}R\_B\}\; =\; \backslash frac\{\backslash frac\{2\}\{3\}\}\{3\}\; =\; \backslash frac\{2\}\{9\}$

Therefore

$P\_\{\backslash text\{parallel\}\}\; :\; P\_\{\backslash text\{series\}\}\; =\; 2\; :\; 9$

### 6. Conclusion

The ratio of the total power outputs for the two configurations (parallel to series) is 2 : 9, exactly matching the provided correct answer.