When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1=2.25 J2 then the value of R in Ω is

P=i2R,i=nER+nr inseries, i=ER+rn inparallel Given that J1 = 2.25 J2 on solving we get R=4Ω

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When two identical batteries of internal resistance $1 Ω$ each are connected in series across a resistor R, the rate of heat produced in R is J₁. When the same batteries are connected in parallel across R, the rate is J₂. If $J_{1} = 2 .25 J_{2}$ then the value of R in $Ω$ is

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answer is 4.

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Detailed Solution

$P = i^{2} R, i = \frac{nE}{R + nr} inseries, i = \frac{E}{R + \frac{r}{n}} inparallel$

Given that J₁ = 2.25 J₂

on solving we get R=4 $Ω$

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