JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 23)
Model a torch battery of length $$l$$ to be made up
of a thin cylindrical bar of radius ‘a’ and a
concentric thin cylindrical shell of radius ‘b’
filled in between with an electrolyte of
resistivity $$\rho $$ (see figure). If the battery is
connected to a resistance of value R, the
maximum Joule heating in R will take place for :
_3rd_September_Morning_Slot_en_23_1.png)
$$R = {{2\rho } \over {\pi l}}\ln \left( {{b \over a}} \right)$$
$$R = {\rho \over {\pi l}}\ln \left( {{b \over a}} \right)$$
$$R = {\rho \over {2\pi l}}\ln \left( {{b \over a}} \right)$$
$$R = {\rho \over {2\pi l}}\left( {{b \over a}} \right)$$
Explanation
_3rd_September_Morning_Slot_en_23_2.png)
Internal resitance of battery
$$\int {dr} = \int\limits_a^b {{{\rho dr} \over {2\pi rl}}} $$
$$ \Rightarrow $$ r = $${{\rho \over {2\pi l}}\ln {b \over a}}$$
Power in external resistance PR = $${\left( {{ \in \over {r + R}}} \right)^2}R$$
Maximum Power in external resistance is generated when it is equal to internal resistance of battery. So PR is maximum when r = R.
$$ \therefore $$ R = $${{\rho \over {2\pi l}}\ln {b \over a}}$$
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