in the circuit shown below C. = 10 uF. c, = = 20 pF, and C = 40 MF. If the charge on C, is 20 C then potential difference between X and Y is C с G HH A X- 30 ro Y HH С. 0020 (2) 3 v (3) 6 V (4) 3.5 V
Answers
Full Adder is the adder which adds three inputs and produces two outputs. The first two inputs are A and B and the third input is an input carry as C-IN. The output carry is designated as C-OUT and the normal output is designated as S which is SUM.
Answer:
: when capacitors are in series connection, the ratio of their capacitance is equal to the ratio of the charge passing through them. This relation helps us find the value of charge through ${C_2}$. We will use this charge value in determining the potential difference across ${C_2}$. Also, the sum of their charges will be equal to the total charge on their positive plates.
As ${C_3}$ and ${C_4}$ are in series, the resultant of their capacitance will be additive. Because charge in a circuit will be constant, the charge found out in the loop of ${C_1}$ and ${C_2}$ is the charge in the loop of ${C_3}$ and ${C_4}$. Thus we will be able to find out the potential difference across ${C_3}$.
Therefore, we will add both the potential difference and find out the given circuit’s potential difference value.
Formulae used: Ratio of their capacitance in ${C_1}$ and ${C_2}$ is equal to the ratio of the charge passing through them:
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{C_1}}}{{{C_2}}}$
Where ${C_1}$ and ${C_2}$ are the capacitance values and is expressed in microFarads$(\mu F)$ and ${Q_1}$ and ${Q_2}$ are the charge values and is expressed in microCoulombs $(\mu C)$.
Sum of their charges: $Q = {Q_1} + {Q_2}$
Where $Q$ is the sum of charge on their positive plates of ${C_1}$ and ${C_2}$.
Potential difference across ${C_2}$: ${V_2} = \dfrac{{{Q_2}}}{{{C_2}}}$
Where ${V_2}$ is the potential difference value across ${C_2}$ and is expressed in volts $(V)$.
Resultant of their capacitance of ${C_3}$ and ${C_4}$: \[{C_3} + {C_4}\]
Where ${C_3}$ and ${C_4}$ are the capacitance values and are expressed in microFarads$(\mu F)$.
Potential difference across ${C_3}$: \[{V_3} = \dfrac{{{Q_1} + {Q_2}}}{{{C_3} + {C_4}}}\]
Where is the potential difference value across ${C_3}$ and is expressed in volts $(V)$.
Given circuit’s potential difference value: $V = {V_2} + {V_3}$
Where $V$is the potential difference values of the entire circuit and is expressed in volts $(V)$.
Complete step by step solution:
From the given circuit we determine that ${C_1}$ & ${C_2}$ and ${C_3}$ & ${C_4}$ are in parallel connection with each other’s loops. And ${C_1}$ & ${C_2}$ and ${C_3}$ & ${C_4}$ are in series among themselves.
Let, ${Q_1},{Q_2},{Q_3},{Q_4}$be the charges on \[{C_1},{C_2},{C_3},{C_4}\].
We know that for capacitors in series connection, the ratio of their capacitance is equal to the ratio of the charge passing through them.
Therefore, $\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{C_1}}}{{{C_2}}}$ .
Substituting the values we get,
\[
\dfrac{{20}}{{{Q_2}}} = \dfrac{{10}}{{20}} \\
\Rightarrow {Q_2} = 40\mu C \\
\]
The potential difference across ${Q_2}$ is $\dfrac{{{Q_2}}}{{{C_2}}}$.
Substituting the values we get,
${V_2} = \dfrac{{{Q_2}}}{{{C_2}}} = \dfrac{{40}}{{20}}$
$ \Rightarrow {V_2} = 2V$
Also, the total charge on the positive plates of ${C_1}$ and ${C_2}$ $Q = {Q_1} + {Q_2}$
Substituting the values we get,
$
Q = 20 + 40 \\
\Rightarrow Q = 60\mu C \\
$
For the second loop of capacitors which are connected in series, the summation of their capacitances is \[{C_3} + {C_4}\].
Substituting the values we find the total capacitance of the loop to be \[20 + 40 = 60\mu F\].
We know that the amount of charge entering a circuit through one end is equal to the amount of charge exiting through the other end.
Therefore we can determine the potential difference across ${C_3}$ to be \[{V_3} = \dfrac{{{Q_1} + {Q_2}}}{{{C_3} + {C_4}}}\].
Substituting the values we get,
\[
{V_3} = \dfrac{{60}}{{60}} \\
\Rightarrow {V_3} = 1V \\
\]
Finally we can calculate the potential difference across $X$ and $Y$ with $V = {V_2} + {V_3}$.
Substituting the values we get the potential difference values of the entire circuit,
$
V = {V_2} + {V_3} = 2V + 1V \\
\Rightarrow V = 1V \\
$
Additional information: capacitors are used to store charge for future requirement of a circuit. Therefore, we add the capacitance value for two capacitors in series connection.
In conclusion, the potential difference between $X$ and $Y$ is $1V$.
Note: The charge is conserved in a given system. Calculating it only for the first loop will be sufficient. Double calculation will lead to error in further calculations involving charge.