Search any question & find its solution
Question:
Answered & Verified by Expert
In the given figure, the capacitors $C_1, C_3, C_4, C_5$ have a capacitance $4 \mu \mathrm{F}$ each. If the capacitor $C_2$ has a capacitance $10 \mu \mathrm{F}$, then effective capacitance between $A$ and $B$ will be
Options:
Solution:
1794 Upvotes
Verified Answer
The correct answer is:
$4 \mu \mathrm{F}$
When a battery is applied across $A$ and $B$, then the points $b$ and $c$ will be at the same potential
$\left(\because C_1=C_4=C_3=C_5=4 \mu \mathrm{F}\right)$
Therefore, no charge flows through $\mathrm{C}_2$. $\mathrm{As}, C_1$ and $C_5$ are in series.
$\therefore$ Their equivalent capacitance,
$C^{\prime}=\frac{C_1 \times C_5}{C_1+C_5}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$
Similarly, $C_4$ and $C_3$ are in series. Therefore, their equivalent capacitance
$C^{\prime \prime}=\frac{C_3 \times C_4}{C_3+C_4}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$
Now, $C^{\prime}$ and $C^{\prime \prime}$ are in parallel. Therefore, effective capacitance between $A$ and $B$
$=C^{\prime}+C^{\prime \prime}=2+2=4 \mu \mathrm{F}$
$\left(\because C_1=C_4=C_3=C_5=4 \mu \mathrm{F}\right)$
Therefore, no charge flows through $\mathrm{C}_2$. $\mathrm{As}, C_1$ and $C_5$ are in series.
$\therefore$ Their equivalent capacitance,
$C^{\prime}=\frac{C_1 \times C_5}{C_1+C_5}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$
Similarly, $C_4$ and $C_3$ are in series. Therefore, their equivalent capacitance
$C^{\prime \prime}=\frac{C_3 \times C_4}{C_3+C_4}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$
Now, $C^{\prime}$ and $C^{\prime \prime}$ are in parallel. Therefore, effective capacitance between $A$ and $B$
$=C^{\prime}+C^{\prime \prime}=2+2=4 \mu \mathrm{F}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.