JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 5)
In the circuit shown, the potential difference between A and B is :
_11th_January_Evening_Slot_en_5_1.png)
6 V
3 V
2 V
1 V
Explanation
Potential difference across AB will be equal to battery equivalent across CD
VAB $$=$$ VCD $$=$$ $${{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}} = {{{1 \over 1} + {2 \over 1} + {3 \over 1}} \over {{1 \over 1} + {1 \over 1} + {1 \over 1}}}$$
$$=$$ $${6 \over 3}$$ $$=$$ 2V
VAB $$=$$ VCD $$=$$ $${{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}} = {{{1 \over 1} + {2 \over 1} + {3 \over 1}} \over {{1 \over 1} + {1 \over 1} + {1 \over 1}}}$$
$$=$$ $${6 \over 3}$$ $$=$$ 2V
Comments (0)
