JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 22)
In the circuit shown, charge on the 5 $$\mu $$F
capacitor is :
_5th_September_Evening_Slot_en_22_1.png)
5.45 $$\mu $$C
18.00 $$\mu $$C
10.90 $$\mu $$C
16.36 $$\mu $$C
Explanation
_5th_September_Evening_Slot_en_22_2.png)
Let potential of point O v0 = 0.
Now, using junction analysis
We can say, q1 + q2 + q3 = 0
$$ \Rightarrow $$ 2(x – 6) + 4(x – 6) + 5(x) = 0
$$ \Rightarrow $$ x = $${{36} \over {11}}$$
q3 = CV = $${{36} \over {11}} \times 5$$ = 16.36 $$\mu $$C
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