• # question_answer A cell of emf 'E' and internal resistance r is connected across a variable resistor R. Plot a graph showing the variation of terminal potential V with resistance R. Predict from the graph, the condition under which V becomes equal to E.   OR Apply Kirchhoff's laws to the loops ACBPA and ACBQA to write the expressions for the currents ${{I}_{1}},{{I}_{2}}$ and ${{I}_{3}}$ in the network given below. Circuit diagram of loops

Answer:

According to given conditions in the question, the circuit diagram can be given as, From the above diagram,             $V=IR=\frac{E}{R+r}\cdot R=\frac{E}{\frac{R+r}{R}}=\frac{E}{1+\frac{r}{R}}$             $\therefore$      $V=\frac{E}{1+\frac{r}{R}}$ When    R = 0, V = 0 When    R = r, $V=\frac{E}{2}$ When    $R=\infty ,$ V = E The plot showing variation of V vs R is given below. From the above graph, it is clear that V becomes equal to E when $R\to \infty$ Or Apply Kirchhoff's 1st law at point A of the circuit             ${{I}_{3}}={{I}_{1}}+{{I}_{2}}$                              ?(i) Applying Kirchhoff's IInd law to loop ACBPA $-12{{I}_{3}}-0.5{{I}_{1}}+6=0$ or $0.5{{I}_{1}}+12{{I}_{3}}=6$ ?(ii) Applying Kirchhoff's IInd law to loop ACBQA             $-12{{I}_{3}}-{{I}_{2}}+10=0$ or,        ${{I}_{2}}+12{{I}_{3}}=10$ The Eqs. (ii) and (iii) are required expressions.

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