JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 29)
The charge accumulated on the capacitor connected in the following circuit is _______ $$\mu \mathrm{C}$$ (Given $$\mathrm{C}=150 \mu \mathrm{F})$$
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Answer
400
Explanation
_27th_January_Morning_Shift_en_29_2.png)
$$\begin{aligned} & \mathrm{V}_{\mathrm{A}}+\frac{10}{3}(1)-6(1)=\mathrm{V}_{\mathrm{B}} \\ & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\frac{10}{3}=\frac{8}{3} \mathrm{volt} \\ & \mathrm{Q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right) \\ & =150 \times \frac{8}{3}=400 \mu \mathrm{C} \end{aligned}$$
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