Test Index

BITSAT 2020 Paper

Section: Physics

Question : 21 of 150

Marks: +1, -0

\mu\text{C},

5\mu\text{F}

Solution:

Potential difference across the branch de is

6\text{V}

Net capacitance of de branch is

2.1\mu\text{F}

So,

q=CV

\Rightarrow q=2.1 \times 6\,\mu\text{C}

\Rightarrow q=12.6\,\mu\text{C}

Potential across

3\mu\text{F}

capacitance is

V=\frac{12.6}{3}=4.2\text{Volt}

Potential across 2 and 5 combination in paralle is

6-4.2=1.8\,\text{V}

So,

q'=(1.8)(5)=9\,\mu\text{C}

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