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When two identical batteries of internal resistance 2 \Omega 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 acros R, the rate is J_2. If 4J1 = 9J2  then value of R in \Omega is:

Option: 1

16\Omega


Option: 2

4\Omega


Option: 3

8\Omega


Option: 4

2\Omega


Answers (1)

best_answer

Let emf of battery be \varepsilon.

In series,

\begin{aligned} & R_{e f f}=(R+4) \Omega \\ & i_1=\frac{2 \varepsilon}{R+4} \\ & J_1=i_1^2 R \\ & =\left(\frac{2 \varepsilon}{R+4}\right)^2 R \end{aligned}

In parallel connection,

\begin{aligned} & R_{\text {eff }}=R+1 \\ & J_2=i_2^2 R=\left(\frac{\varepsilon}{R+1}\right)^2 R \\ & \because 4 J_1=9 J_2 \\ & 4\left(\frac{2 \varepsilon}{R+4}\right)^2 R=9\left(\frac{\varepsilon}{R+1}\right)^2 R \end{aligned}

Taking square root,

\begin{aligned} & 2 \times \frac{2 \varepsilon}{R+4}=\frac{3 \varepsilon}{R+1} \\ & \Rightarrow 4 R+4=3 R+12 \\ & \Rightarrow 4 R-3 R=12-4 \\ & \Rightarrow R=8 \Omega \end{aligned}

Posted by

Gautam harsolia

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