When two resistances $R_{1}\ and\ R_{2}$ are connected in series, they consume 12 W powers. When they are connected in parallel, they consume 50 W powers. What the ratio of the powers of $R_{1}\ and\ R_{2}?$

Question

Dr. Abubacker · Accepted Answer

(c) In series,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{s} = \frac{P_{1}\  	imes P_{2}}{P_{1} + P_{2}} = 12\ W$
In parallel, $P_{p} = P_{1} + PP_{2} = 50\ W$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{1}P_{2} = 12 	imes 50 = 600$
Now, $\left( P_{1} - P_{2} ight)^{2} = \left( P_{1} + P_{2} ight)^{2} - 4P_{1}P_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = (50)^{2} - 4 	imes 600$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 2500 - 2400 = 100$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ P_{1} - P_{2} = 10$
Or $P_{1} = 30\ W,\ P_{2} = 20\ W$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{P_{1}}{P_{2}} = \frac{3}{2}$

Q. When two resistances $R_{1}\ and\ R_{2}$ are connected in series, they consume 12 W powers. When they are connected in parallel, they consume 50 W powers. What the ratio of the powers of $R_{1}\ and\ R_{2}?$

Explanation

Reviewed by Dr. Abubacker

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