JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 12)
In the given circuit diagram, a wire is joining
points B and D. The current in this wire is :
_9th_January_Morning_Slot_en_12_1.png)
_9th_January_Morning_Slot_en_12_1.png)
4 A
2 A
zero
0.4 A
Explanation
_9th_January_Morning_Slot_en_12_2.png)
_9th_January_Morning_Slot_en_12_3.png)
_9th_January_Morning_Slot_en_12_4.png)
I = $${{20} \over 2}$$ = 10 A
_9th_January_Morning_Slot_en_12_5.png)
I1 = $${4 \over 5}I$$, I2 = $${1 \over 5}I$$
$$ \Rightarrow $$ I1 = 8A, I2 = 2 A
I1' = $${3 \over 5} \times 10$$ = 6 A
I2' = $${2 \over 5} \times 10$$ = 4 A
$$ \Rightarrow $$ IBC = 8 – 6 = 2 A
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