Question

In the circuit, the potential difference between A and B is

1.1/10 V
2.2.5V
3.10V
4.5V​

Answer 1

The potential difference between A and B is 2 V.

Explanation:

According to figure,

Here, C₃ and C₄  are connected in series.

We need to calculate the capacitance

Using formula of capacitance

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

Put the value into the formula

$\dfrac{1}{C}=\dfrac{1}{8}+\dfrac{1}{8}$

$C=4\ \mu F$

Now, C and C₂ are connected in parallel

Then the capacitance is

$C'=C+C₂$

Put the value into the formula

$C'=4+4$

$C'=8\ \mu F$

Now, C₁ and C' are connected in series

Then, the capacitance is

$\dfrac{1}{C''}= \dfrac{1}{C_{1}}+\dfrac{1}{C'}$

Put the value into the formula

$\dfrac{1}{C''}=\dfrac{1}{2}+\dfrac{1}{8}$

$C''=1.6\ \mu F$

We need to calculate the charge

Using formula of charge

$Q=CV$

Put the value into the formula

$Q=1.6\times20$

$Q=32\ \mu C$

We know that,

Charge is same in series.

We need to calculate the charge on 8 μF

Here, C₁ and C₂ are connected in series so the charge will be same 32 μC.

C₂ and C in parallel and equal in magnitude so the charge will be divided 16 μC.

C₃ and C₄ in series so the charge will be same 16 μC.

We need to calculate the potential difference between A and B

Using formula of voltage

$V=\dfrac{Q}{C}$

$V=\dfrac{16}{8}$

$V=2\ V$

Hence, The potential difference between A and B is 2 V.

Learn more :

Topic : capacitance

https://brainly.in/question/11103154

Answer 2

Answer:Check out the attachment.

Explanation:How to deal with such questions:

Step 1: Find the effective capacitance.

Step 2: Find Q on different elements if they’re in parallel. If they’re in series, find V on different elements. Note: In series connection, Q is same for all elements in series. In parallel connection, V across all the elements are same.

Step 3: You would have reduced the circuit for finding equivalent capacitance. Now, elaborate the circuit again and find either Q or V (accordingly) across each individual capacitors.

Step 4: Use C=Q/V for individual capacitors for finding either the charge stored on the capacitor under consideration or the potential difference across it.