Question

⚡Two resistors R1 = 400 ohm and R2 = 20 ohm are connected in parallel to a battery.If heating power developed in R1 is 25W find the heating power developed in R2.​

Answer 1

Given,

R1 = 400 ohm.

R2 = 20ohm.

Power developed in R1 = P1 = 25 W

Let the power developed in R2 be P2.

As per the question,

The resistance are in parallel combination, thus the potential difference (V) across R1 and R2 will be same.

we know,

$V^{2}=PR$

$V^{2}= P1R1 \\\\V^{2}= P2R2$

∴ P1R1 = P2R2

on putting the values,

25*400 = P2 *  20

P2 = 500 W.

∴ The power developed in R2 is 500 W.

Answer 2

$\huge{\bold{\boxed{\boxed{\mathfrak{\red{ANSWER}}}}}}$

★ $\bold{500w}$

$\large{\underline{\underline{\mathbf{Explanation:-}}}}$

$\star{\mathtt{\blue{\underline{Given:-}}}}$

$\large\bold{Resistance\:R_{1}=400\:ohm}$

$\large\bold{Resistance\:R_{2}=20\:ohm}$

$\bold{Power\:in\:resistor,R_{1}, P_{1} =25\:W}$

$\bold{Using\:the\:Formula,}$

$\bold{\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }}$

$\large\bold{=>\frac{25}{P_{2} } =\frac{20}{400}}$

$\large\bold{=>P_{2}=500\:W}$

$\bold{This\:is\:the\:heating\:power\:developed\:in\:R_{2}}$

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