Six identical capacitors with capacitance C are connected as shown in the figure (Figure 1) ….
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Question “Six identical capacitors with capacitance C are connected as shown in the figure (Figure 1) ….”
Six identical capacitors with capacitance
C
are connected as shown in the figure
(Figure 1)
.
1)What
is the potential difference between points a and b?
2)
What is the equivalent capacitance of these six
capacitors?
Answer
This problem can be solved using the series and parallel combination of capacitance and the charge across capacitance concepts.
First, calculate the equivalent capacitance by using the series or parallel combination of the capacitor. Next, calculate the charge across circuit using the expression of charge, which is a term that includes capacitance and potential.
Capacitance refers to the conductor’s ability to store an electrical charge. Farad is the SI unit for capacitance.
If N
If N \frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}
The voltage division rule gives the voltage of the capacitor:
{V_X} = {V_S}\frac{{{C_T}}}{{{C_X}}}Here, {C_T}
This is the equivalent circuit to the one shown.
(a)
At point a, the capacitors are identical
{V_a} = VThe voltage is also dropped when it reaches b {V_b} = V
Potential at point $math_tag_5
\begin{array}{c}\\{V_a} - {V_b} = 0\\\\\Delta V = 0\\\end{array}[Part a]
Part of
Part of
(b)
If N \frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}..... + \frac{1}{{{C_N}}}
Two capacitors are shown in the diagram as part A.
\begin{array}{c}\\\frac{1}{{{C_{equ}}}} = \frac{1}{C} + \frac{1}{C}\\\\ = \frac{2}{C}\\\\{C_{equ}} = \frac{C}{2}\\\end{array}If N
In the diagram, part B, you can see three capacitors connected in parallel.
\begin{array}{c}\\{C_{equ}} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2}\\\\ = \frac{{3C}}{2}\\\end{array}[Part b]
Part B
Part B
Ans: Part A
The difference in points aPart
The equivalent capacitance for the six capacitors is $math_tag_17
Conclusion
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