Two resister $R_{1} = 400 Ω$ and $R_{2} = 20 Ω$ are connected in parallel to a battery. If heating power developed in $R_{1}$ in 25 W, find the heating power developed in $R_{2}$ .

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Solution

Heat $= i^{2} R t$
Let current through the resistance $R_{1}$ be $i_{1}$ and the voltage across it be $V_{1}$ .
$25 = i_{1}^{2} \times 400 \times 1$
$i_{1} = 0.25 A$
$V_{1} = i_{1} R_{1}$
$= 0.25 \times 400$
$= 100 V$
Let current through the resistance $R_{2}$ be $i_{2}$ and the voltage across it be $V_{2}$ .
$V_{2} = i_{2} R_{2}$
$100 = i_{2} \times 20$
$i_{2} = 5 A$
So, Heating power $= i_{2}^{2} R_{2} t$
$= 5^{2} \times 20 \times 1$
$= 500 W$

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Two resister R1=400Ω and R2=20Ω are connected in parallel to a battery. If heating power developed in R1 in 25 W, find the heating power developed in R2.

Heat=i2RtLet current through the resistance R1 be i1 and the voltage across it be V1.25=i21×400×1i1=0.25AV1=i1R1=0.25×400=100VLet current through the resistance R2 be i2 and the voltage across it be V2.V2=i2R2100=i2×20i2=5ASo, Heating power =i22R2t =52×20×1 =500W

Two resister $R_{1} = 400 Ω$ and $R_{2} = 20 Ω$ are connected in parallel to a battery. If heating power developed in $R_{1}$ in 25 W, find the heating power developed in $R_{2}$ .