JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Graphical Questions

  • question_answer
    Battery shown in figure has e.m.f. E and internal resistance r. Current in the circuit can be varied by sliding the contact J. If at any instant current flowing through the circuit is I, potential difference between terminals of the cell is V, thermal power generated in the cell is equal to h fraction of total electrical power generated in it.; then which of the following graph is correct

    A)                                  

    B)

    C)                                        

    D)            Both (a) and (b) are correct

    Correct Answer: D

    Solution :

                       Terminal voltage \[V=E-Ir.\] Hence the graph between V and i will be a straight line having negative slope and positive intercept. Thermal power generated in the external circuit \[P=EI-{{I}^{2}}r.\] Hence graph between P and I will be a parabola passing through origin. Also at an instant, thermal power generated in the cell = \[{{i}^{2}}r\] and total electrical power generated in the cell = Ei. Hence the fraction \[\eta =\frac{{{I}^{2}}r}{EI}=\left( \frac{r}{E} \right)\,I;\] so \[\eta \propto I\]. It means graph between h and I will be a straight line passing through origin.


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