What is the equivalent capacitance of the three capacitors in Figure P23.36? 36. Il What is…
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Question “What is the equivalent capacitance of the three capacitors in Figure P23.36? 36. Il What is…”
What is the equivalent capacitance of the three capacitors in
Figure P23.36?
Answer
Answer 1
This question is based upon the idea of equivalent capacitance between series and parallel combinations of capacitors.
Calculate the equivalent capacitance for the first arm, then calculate the equivalent capacitance for both parallel capacitors for both arms.
Capacitor refers to the device that stores energy from the electric field. Capacitance refers to the property of capacitors that store the charge. Farad is its unit.
A capacitor with equivalent capacitance can be used to represent two or more capacitors in an electrical circuit. This is how the equivalent capacitance can look like:
\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{{C_1}}} + {\frac{1}{C}_2} + ...\frac{1}{{{C_{\rm{n}}}}}Here $math_tag_1
A capacitor with equivalent capacitance can be used to represent two or more capacitors in an electrical circuit. This is how you can represent the equivalent capacitance.
{C_{{\rm{eq}}}} = {C_1} + {C_2} + {C_3}{\rm{ }}... + {C_{\rm{n}}}Here $math_tag_1
(a)
The circuit shown is as follows:
The equivalent capacitance for the first arm can be calculated as follows:
Substitute $math_tag_4
\begin{array}{c}\\\frac{1}{{{C_{12}}}} = \frac{1}{{20{\rm{ }}\mu {\rm{F}}}} + \frac{1}{{30{\rm{ }}\mu {\rm{F}}}}\\\\\frac{1}{{{C_{12}}}} = \frac{{\left( {30{\rm{ }}\mu {\rm{F}}} \right) + \left( {20{\rm{ }}\mu {\rm{F}}} \right)}}{{\left( {20{\rm{ }}\mu {\rm{F}}} \right)\left( {30{\rm{ }}\mu {\rm{F}}} \right)}}\\\\{C_{12}} = 12{\rm{ }}\mu {\rm{F}}\\\end{array}Therefore, the equivalent capacitance for first arm is $math_tag_6
The equivalent capacitance for the circuit can be calculated as follows:
Substitute $math_tag_6
\begin{array}{c}\\{C_{eq}} = 12{\rm{ }}\mu {\rm{F}} + 25{\rm{ }}\mu {\rm{F}}\\\\ = 37{\rm{ }}\mu {\rm{F}}\\\end{array}The equivalent capacitance for the circuit is therefore $math_tag_9
(b)
The circuit shown is as follows:
The equivalent capacitance for the first arm can be calculated as follows:
Substitute $math_tag_4
\begin{array}{c}\\{C_{12}} = 20{\rm{ }}\mu {\rm{F}} + 60{\rm{ }}\mu {\rm{F}}\\\\ = 80{\rm{ }}\mu {\rm{F}}\\\end{array}Therefore, the equivalent capacitance for first loop is $math_tag_12
The equivalent capacitance for the circuit can be calculated as follows:
Substitute $math_tag_12
\begin{array}{c}\\\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{80{\rm{ }}\mu {\rm{F}}}} + \frac{1}{{10{\rm{ }}\mu {\rm{F}}}}\\\\ = \frac{9}{{80{\rm{ }}\mu {\rm{F}}}}\\\\{C_{{\rm{eq}}}} = 8.89{\rm{ }}\mu {\rm{F}}\\\end{array}The equivalent capacitance of the circuit therefore is $math_tag_15. Ans: Part a
Part b: The equivalent capacitance for the three capacitors is $math_tag_9
The equivalent capacitance for the three capacitors is $math_tag_15
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