Four capacitors ${C_1}$, ${C_2}$, ${C_3}$, ${C_4}$ are connected as shown in figure below. Calculate the equivalent capacitance of the circuit between points X and Y.
Answer
653.1k+ views
Hint: Use the formula for equivalent capacitance when the capacitors are connected in series and parallel. Remember that in a set of parallel capacitors the total capacitance is just a sum of the individual capacitors and in the series capacitance, the reciprocal sum of the individual capacitors gives the total capacitance.
Formula used:
\[\dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}\]
Where \[{C'_1},{C_2},{C_3}\] are capacitance of the capacitance.
Complete step by step answer:
We know that,
\[ \Rightarrow \dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}\]
Therefore,
\[ \Rightarrow \dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{20}} + \dfrac{1}{{30}}\]
Therefore,
\[ \Rightarrow {C'_1} = 12\mu F\]
Since the capacitors \[{C_1}^{'}\] and \[{C_4}\] are connected in series,
\[ \Rightarrow {C_1}’’ = 12 + 28\]
\[ \Rightarrow {C_1}‘’ = 40\mu F\]
From this we can calculate,
\[ \Rightarrow \dfrac{1}{C} = \dfrac{1}{{10}} + \dfrac{1}{{40}}\]
\[ \Rightarrow \dfrac{{4 + 1}}{{40}}\]
\[ \Rightarrow \dfrac{1}{C} = \dfrac{5}{{40}}\]
\[\therefore \dfrac{1}{8}\]
Therefore,
\[C = 8\mu F\]
Thus, we can conclude that the equivalent capacitance of the circuit between points X and Y is 8 F.
Additional information:
Capacitors can also be called electric condensers. It is an electrical component with two terminals, that can store energy in the form of electrical charge. Capacitors are usually used to increase the effects of capacitance. Capacitance is a proportional constant that relates charge to the potential difference between two conductors. It is used to measure in the units of Farad.
We know,
\[Q = \dfrac{C}{V}\]
Where $C$=capacitance, $Q$= charge contained in the capacitor, $V$= potential difference between two conductors.
Usually, most capacitors have two electrical conductors separated by a metallic plate, also known as the dielectric medium. Most capacitors are used in appliances that require sudden huge bursts of energy, like fans, air conditioners, mobile phones. A capacitor is always required to be placed in a power circuit with a given applied voltage of either alternating or direct current. Capacitors are used to separate the A.C and D.C, to filter out the unwanted signals, used as a sensing device and used in the audio system of the vehicle.
Note:
If the capacitors, ${C_1}$, ${C_2}$ are connected in series, the equivalent capacitance is given by $\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}...$. On the other hand, if the capacitors, ${C_1}$, ${C_2}$ are connected in parallel then the equivalent capacitance is given by $C = {C_1} + {C_2}...$
Formula used:
\[\dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}\]
Where \[{C'_1},{C_2},{C_3}\] are capacitance of the capacitance.
Complete step by step answer:
We know that,
\[ \Rightarrow \dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}\]
Therefore,
\[ \Rightarrow \dfrac{1}{{{{C'}_1}}} = \dfrac{1}{{20}} + \dfrac{1}{{30}}\]
Therefore,
\[ \Rightarrow {C'_1} = 12\mu F\]
Since the capacitors \[{C_1}^{'}\] and \[{C_4}\] are connected in series,
\[ \Rightarrow {C_1}’’ = 12 + 28\]
\[ \Rightarrow {C_1}‘’ = 40\mu F\]
From this we can calculate,
\[ \Rightarrow \dfrac{1}{C} = \dfrac{1}{{10}} + \dfrac{1}{{40}}\]
\[ \Rightarrow \dfrac{{4 + 1}}{{40}}\]
\[ \Rightarrow \dfrac{1}{C} = \dfrac{5}{{40}}\]
\[\therefore \dfrac{1}{8}\]
Therefore,
\[C = 8\mu F\]
Thus, we can conclude that the equivalent capacitance of the circuit between points X and Y is 8 F.
Additional information:
Capacitors can also be called electric condensers. It is an electrical component with two terminals, that can store energy in the form of electrical charge. Capacitors are usually used to increase the effects of capacitance. Capacitance is a proportional constant that relates charge to the potential difference between two conductors. It is used to measure in the units of Farad.
We know,
\[Q = \dfrac{C}{V}\]
Where $C$=capacitance, $Q$= charge contained in the capacitor, $V$= potential difference between two conductors.
Usually, most capacitors have two electrical conductors separated by a metallic plate, also known as the dielectric medium. Most capacitors are used in appliances that require sudden huge bursts of energy, like fans, air conditioners, mobile phones. A capacitor is always required to be placed in a power circuit with a given applied voltage of either alternating or direct current. Capacitors are used to separate the A.C and D.C, to filter out the unwanted signals, used as a sensing device and used in the audio system of the vehicle.
Note:
If the capacitors, ${C_1}$, ${C_2}$ are connected in series, the equivalent capacitance is given by $\dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}...$. On the other hand, if the capacitors, ${C_1}$, ${C_2}$ are connected in parallel then the equivalent capacitance is given by $C = {C_1} + {C_2}...$
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

A member of Simon commission later became Prime Minister class 12 social science CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Give one example of a reptile that is viviparous class 12 biology CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

