In the given circuit each of the resistors is of $1 \mathrm{k} \Omega$ resistance and each of the ca...

Once a capacitor is fully charged, it behaves like an open circuit for DC current. So, we can remove all the capacitors to simplify the circuit and then obtain potential difference across $B F$.On simplifying we get the following circuit which can be remade as the following:$$ G H \text { will short-circuit the three resistors and the potential at } F \text { will be equal to } G \text {. } $$The current can be calculated using Ohm's law:$$ \begin{gathered} R_{n e t}=4 \times 1 \mathrm{k} \Omega \\ V=I R \\ I=\frac{5}{4} \times 10^{-3} \mathrm{~A} \end{gathered} $$The potential difference between $F$ and $B$ is the same as the potential difference between $H$ and $B$, where the net resistance between the two points is $2 k \Omega$. Hence, the potential difference is:$$ V_{H B}=I\left(R_{H B}\right) \Rightarrow V_{H B}=\frac{5}{4} \times 10^{-3} \times 2 \times 10^3=\frac{10}{4} V $$Using the formula $Q=C V$ :$$ \begin{gathered} Q=4 \times 10^{-6} \cdot \frac{10}{4} \\ Q=10 \mu C \end{gathered} $$

What is the potential difference between the points $P$ and $Q$ in the circuit shown below, once the...

Step 1: Understand the networkThe four capacitors form adiamond (bridge) networkbetween the left and right terminals of a12 V battery.Top-left capacitor =4 μFTop-right capacitor =2 μFBottom-left capacitor =2 μFBottom-right capacitor =4 μFPointPis the junction of the top capacitors.PointQis the junction of the bottom capacitors.Step 2: Voltage division in each branchEach branch behaves like aseries combination:Top branch (4 μF and 2 μF in series)✅Answer:4V ✔️

Capacitor - IAT-IISER Physics PYQs

Once a capacitor is fully charged, it behaves like an open circuit for DC current. So, we can remove all the capacitors to simplify the circuit and then obtain potential difference across

B%20F

.On simplifying we get the following circuit which can be remade as the following:$

G%20H%20%5Ctext%20%7B%20will%20short-circuit%20the%20three%20resistors%20and%20the%20potential%20at%20%7D%20F%20%5Ctext%20%7B%20will%20be%20equal%20to%20%7D%20G%20%5Ctext%20%7B.%20%7D

The%20current%20can%20be%20calculated%20using%20Ohm's%20law%3A

%5Cbegin%7Bgathered%7D%20R_%7Bn%20e%20t%7D%3D4%20%5Ctimes%201%20%5Cmathrm%7Bk%7D%20%5COmega%20%5C%5C%20V%3DI%20R%20%5C%5C%20I%3D%5Cfrac%7B5%7D%7B4%7D%20%5Ctimes%2010%5E%7B-3%7D%20%5Cmathrm%7B~A%7D%20%5Cend%7Bgathered%7D

The%20potential%20difference%20between

and

is%20the%20same%20as%20the%20potential%20difference%20between

and

%2C%20where%20the%20net%20resistance%20between%20the%20two%20points%20is

2 k \Omega

.%20Hence%2C%20the%20potential%20difference%20is%3A

V_%7BH%20B%7D%3DI%5Cleft(R_%7BH%20B%7D%5Cright)%20%5CRightarrow%20V_%7BH%20B%7D%3D%5Cfrac%7B5%7D%7B4%7D%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%202%20%5Ctimes%2010%5E3%3D%5Cfrac%7B10%7D%7B4%7D%20V

Using%20the%20formula

Q=C V

%3A

%5Cbegin%7Bgathered%7D%20Q%3D4%20%5Ctimes%2010%5E%7B-6%7D%20%5Ccdot%20%5Cfrac%7B10%7D%7B4%7D%20%5C%5C%20Q%3D10%20%5Cmu%20C%20%5Cend%7Bgathered%7D

To analyze this circuit, let's start by understanding the initial scenario. Capacitor

C_1

with a capacitance of

2%20%5C%2C%20%5Cmu%20%5Cmathrm%7BF%7D

is connected to a battery with a potential

V

. The charge

Q

C_1

is calculated as:

Q%20%3D%20C_1%20%5Ctimes%20V

After

C_1

is disconnected from the battery, it is connected in parallel with another capacitor,

C_2

, which has a capacitance of

8%20%5C%2C%20%5Cmu%20%5Cmathrm%7BF%7D

. Since they are connected in parallel, they will share the charge while maintaining the same voltage across them. The total charge initially on

C_1

is conserved and distributed between both capacitors.The initial total charge:

Q%20%3D%20C_1%20%5Ctimes%20V

After the connection, let

Q_1

and

Q_2

denote the charges on

C_1

and

C_2

, respectively. In a parallel setup, the voltage across both capacitors is the same, denoted as

V_f

%5Cbegin%7Balign*%7D%20Q_1%20%26%3D%20C_1%20%5Ctimes%20V_f%20%5C%5C%20Q_2%20%26%3D%20C_2%20%5Ctimes%20V_f%20%5Cend%7Balign*%7D

With charge conservation, the equation is:

Q_1%20%2B%20Q_2%20%3D%20Q

Substituting the expressions for

Q_1

and

Q_2

C_1%20%5Ctimes%20V_f%20%2B%20C_2%20%5Ctimes%20V_f%20%3D%20C_1%20%5Ctimes%20V

Solving for

V_f

V_f%20%3D%20%5Cfrac%7BC_1%7D%7BC_1%20%2B%20C_2%7D%20%5Ctimes%20V

Now, to find the ratio

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D

, we calculate

Q_2

using

V_f

Q_2%20%3D%20C_2%20%5Ctimes%20V_f%20%3D%20C_2%20%5Ctimes%20%5Cfrac%7BC_1%20%5Ctimes%20V%7D%7BC_1%20%2B%20C_2%7D

The total charge is maintained as:

Q_1%20%2B%20Q_2%20%3D%20Q%20%3D%20C_1%20%5Ctimes%20V

Thus, the ratio

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D

becomes:

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D%20%3D%20%5Cfrac%7BC_2%20%5Ctimes%20V_f%7D%7BC_1%20%5Ctimes%20V%7D%20%3D%20%5Cfrac%7BC_2%20%5Ctimes%20%5Cfrac%7BC_1%20%5Ctimes%20V%7D%7BC_1%20%2B%20C_2%7D%7D%7BC_1%20%5Ctimes%20V%7D

This simplifies to:

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D%20%3D%20%5Cfrac%7BC_2%7D%7BC_1%20%2B%20C_2%7D

Substituting the given capacitance values:

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D%20%3D%20%5Cfrac%7B8%20%5C%2C%20%5Cmu%20%5Cmathrm%7BF%7D%7D%7B2%20%5C%2C%20%5Cmu%20%5Cmathrm%7BF%7D%20%2B%208%20%5C%2C%20%5Cmu%20%5Cmathrm%7BF%7D%7D%20%3D%20%5Cfrac%7B8%7D%7B10%7D

Therefore, the ratio is:

%5Cfrac%7BQ_2%7D%7BQ_1%20%2B%20Q_2%7D%20%3D%20%5Cfrac%7B4%7D%7B5%7D

Capacitor

High-Yield Trend

Questions
3 MCQs

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Capacitor

High-Yield Trend

Questions 3 MCQs

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Official Solution

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