Answer
Verified
451.5k+ views
Hint: In this question use the relationship between the capacitance, charge and potential on a capacitor that is $Q = CV$. Use the constraints given in the question to find the charge onto the two capacitors. Then use the fact that since potential is found to be zero therefore the capacitors must be connected in phase opposition (i.e. positive sign connected to positive and negative sign connected to negative) that is ${Q_1} - {Q_2} = 0$. This will help get the right answer to this problem statement.
Complete step-by-step solution:
As we know that the charge stored on the capacitor is the product of individual capacitance and the voltage stored in the capacitor.
$ \Rightarrow Q = CV$.................. (1)
Where, Q = charge, C = capacitance and V = voltage.
Now let the capacitance of first capacitor be ${C_1}$ and the capacitance of second capacitor be ${C_2}$
Charged voltages of the first and second capacitor are 120V and 200V.
So the charges on the first and second capacitor are
$ \Rightarrow {Q_1} = 120{C_1}$ Coulomb
And
$ \Rightarrow {Q_2} = 200{C_2}$Coulomb.
Now it is given that they are connected in parallel and it is found that the potential on each of them is zero.
So from equation (1) if the potential or voltage on each of them is zero then the charge on each of them will be zero.
It is only possible when they connected in phase opposition (i.e. positive sign connected to positive and negative sign connected to negative) as shown in figure, such that ${Q_1} - {Q_2} = 0$
Now substitute the values in above equation we have,
$ \Rightarrow {Q_1} - {Q_2} = 0$
$ \Rightarrow 120{C_1} - 200{C_2} = 0$
Now simplify this we have,
$ \Rightarrow 120{C_1} = 200{C_2}$
Divide by 40 throughout we have,
$ \Rightarrow 3{C_1} = 5{C_2}$
So this is the required relation between ${C_1}$ and ${C_2}$ i.e. first and second capacitor.
Hence option (C) is the correct answer.
Note – The parallel combinatory of the capacitors is exactly equivalent to that series connection of the resistors that is ${C_{eq}} = {C_1} + {C_2}$, it is advised to remember the direct relationship that is $q = cv$. A parallel plate capacitor is formed by charge parallel plates which may have dielectric or even air filled in between the plates. These plates are separated by some distance d and if A is the area of cross-section of the plates then capacitance C is given by $C = \dfrac{{{ \in _0}A}}{d}$. This information is quite useful while dealing with problems involving capacitors.
Complete step-by-step solution:
As we know that the charge stored on the capacitor is the product of individual capacitance and the voltage stored in the capacitor.
$ \Rightarrow Q = CV$.................. (1)
Where, Q = charge, C = capacitance and V = voltage.
Now let the capacitance of first capacitor be ${C_1}$ and the capacitance of second capacitor be ${C_2}$
Charged voltages of the first and second capacitor are 120V and 200V.
So the charges on the first and second capacitor are
$ \Rightarrow {Q_1} = 120{C_1}$ Coulomb
And
$ \Rightarrow {Q_2} = 200{C_2}$Coulomb.
Now it is given that they are connected in parallel and it is found that the potential on each of them is zero.
So from equation (1) if the potential or voltage on each of them is zero then the charge on each of them will be zero.
It is only possible when they connected in phase opposition (i.e. positive sign connected to positive and negative sign connected to negative) as shown in figure, such that ${Q_1} - {Q_2} = 0$
Now substitute the values in above equation we have,
$ \Rightarrow {Q_1} - {Q_2} = 0$
$ \Rightarrow 120{C_1} - 200{C_2} = 0$
Now simplify this we have,
$ \Rightarrow 120{C_1} = 200{C_2}$
Divide by 40 throughout we have,
$ \Rightarrow 3{C_1} = 5{C_2}$
So this is the required relation between ${C_1}$ and ${C_2}$ i.e. first and second capacitor.
Hence option (C) is the correct answer.
Note – The parallel combinatory of the capacitors is exactly equivalent to that series connection of the resistors that is ${C_{eq}} = {C_1} + {C_2}$, it is advised to remember the direct relationship that is $q = cv$. A parallel plate capacitor is formed by charge parallel plates which may have dielectric or even air filled in between the plates. These plates are separated by some distance d and if A is the area of cross-section of the plates then capacitance C is given by $C = \dfrac{{{ \in _0}A}}{d}$. This information is quite useful while dealing with problems involving capacitors.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Which are the Top 10 Largest Countries of the World?
Difference Between Plant Cell and Animal Cell
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE