SATHEE: Current Electricity 5 Question 15

Current Electricity 5 Question 15

15. 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 $J_{1}$ . When the same batteries are connected in parallel across $R$ , the rate is $J_{2}$ . If $J_{1} = 2.25 J_{2}$ then the value of $R$ in $Ω$ is

(2010)

Analytical & Desrcriptive Questions

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Solution:

In series, $i = \frac{2 E}{2 + R} \Rightarrow J_{1} = i^{2} R = {\frac{2 E}{2 + R}}^{2} \cdot R$

In parallel, $i = \frac{E}{0.5 + R}$

$\begin{aligned} ∴ J_{2} & = i^{2} R = \frac{E}{0.5 + R}^{2} \cdot R \\ \frac{J_{1}}{J_{2}} & = 2.25 \\ = \frac{4 (0.5 + R)^{2}}{(2 + R)^{2}} or 1.5 = \frac{2 (0.5 + R)}{(2 + R)} \end{aligned}$

Solving we get, $R = 4 Ω$

$∴$ The answer is 4 .

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Current Electricity 5 Question 15

15. When two identical batteries of internal resistance 1Ω 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=2.25J2 then the value of R in Ω is

Analytical & Desrcriptive Questions

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