Two heaters A and B have power ratings of 1 t | vedclass

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(A) The power rating $P$ of a heater is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since $V$ is constant,$R = \fr

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$2: 9$

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(A) The power rating $P$ of a heater is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since $V$ is constant,$R = \frac{V^2}{P}$.For heater $A$,$P_A = 1 	ext{ kW}$,so $R_A = \frac{V^2}{1}$.For heater $B$,$P_B = 2 	ext{ kW}$,so $R_B = \frac{V^2}{2}$.Thus,$R_A = 2R_B$.In series combination,the equivalent resistance is $R_S = R_A + R_B = 2R_B + R_B = 3R_B$. The power consumed is $P_S = \frac{V^2}{R_S} = \frac{V^2}{3R_B}$.In parallel combination,the equivalent resistance is $R_P = \frac{R_A R_B}{R_A + R_B} = \frac{(2R_B)(R_B)}{2R_B + R_B} = \frac{2R_B^2}{3R_B} = \frac{2}{3}R_B$. The power consumed is $P_P = \frac{V^2}{R_P} = \frac{V^2}{(2/3)R_B} = \frac{3V^2}{2R_B}$.The ratio of power outputs is $\frac{P_S}{P_P} = \frac{V^2 / 3R_B}{3V^2 / 2R_B} = \frac{1}{3} 	imes \frac{2}{3} = \frac{2}{9}$.

Two heaters $A$ and $B$ have power ratings of $1 \text{ kW}$ and $2 \text{ kW}$,respectively. These 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:

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