Question

the equivalent capacitance between the points A and B in the given diagram is​

Answer 1

ans- is (c)8/3uF,hope it help u,pls mark me as brainlist

Answer 2

The equivalent capacitance between the points A and B in the given diagram is​ $\dfrac{8}{3}\ \mu F$.

Explanation:

It is given that,

Capacitance of all the capacitors, $C_1=C_2=C_3=C_4=2\ \mu F$

The three capacitors that are placed in the lower part of the circuit are in series. The equivalent capacitance is given by :

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

$\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}$

$C = \dfrac{2}{3}\ \mu F$

Now C and the capacitor that is placed between A and B are in parallel. Equivalent capacitance is given by :

$C_{eq}=C_4+C$

$C_{eq}=\dfrac{2}{3}+2$

$C_{eq}=\dfrac{8}{3}\ \mu F$

So, the equivalent capacitance between the points A and B in the given diagram is​ $\dfrac{8}{3}\ \mu F$. Hence, this is the required solution.

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