Question 58 of 180

The equivalent capacitance between $A$ and $B$ in the circuit given below is:

A$4.9 \mu \mathrm{F}$B$3.6 \mu \mathrm{F}$C$5.4 \mu \mathrm{F}$D$2.4 \mu \mathrm{F}$

Show answer and solution▾

Answer: (D) $2.4 \mu \mathrm{F}$

The simplified circuit of the circuit given in question as follows:


The equivalent capacitance between C \& D capacitors of $2 \mu \mathrm{F}, 5 \mu \mathrm{F}$ and $5 \mu \mathrm{F}$ are in parallel.
$\therefore \quad \mathrm{C}_{\mathrm{CD}}=2+5+5=12 \mu \mathrm{F}(\because$ In parallel grouping $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2+\ldots .+\mathrm{C}_{\mathrm{n}}$ )
Similarly equivalent capacitance between $\mathrm{E}$ $\& \mathrm{BC}_{\mathrm{EB}}$ $=4+2=6 \mu \mathrm{F}$
Now equivalent capacitance between A \& B $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{6}+\frac{1}{12}+\frac{1}{6}=\frac{5}{12}$
$\Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{12}{5}=2.4 \mu \mathrm{F}(\because$ In series grouping, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\ldots \ldots \ldots+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ )

JEE Main Physics practice question on Capacitors (Grouping of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.