Calculate the equivalent capacitance between points a and b for each of the two networks shown in the figure below. Each capacitor has a capacitance of 5.22 µF. (Give your answers to at least two decimal places.)
Network 1 µF
Network 2 µF
www.webassign.net/katzpse1/27-p-021.png
To calculate the equivalent capacitance between points a and b for each network, we need to analyze the circuit and use the appropriate formulas. Let's examine each network separately.
Network 1:
In this network, the two capacitors are connected in parallel, as they share the same endpoints (points a and b). The formula for calculating the equivalent capacitance for capacitors in parallel is:
1/C_eq = 1/C_1 + 1/C_2
where C_eq is the equivalent capacitance and C_1, C_2 are the capacitances of the individual capacitors.
Substituting the given values:
1/C_eq = 1/5.22 µF + 1/5.22 µF
To simplify the equation, we can combine the fractions:
1/C_eq = 2/5.22 µF
To find the inverse of this expression, we flip both sides of the equation:
C_eq = 5.22 µF/2
Calculating this:
C_eq = 2.61 µF
Therefore, the equivalent capacitance between points a and b in Network 1 is 2.61 µF.
Network 2:
In this network, the two capacitors are connected in series, meaning their total capacitance is the sum of their individual capacitances. The formula for calculating the equivalent capacitance for capacitors in series is:
C_eq = C_1 + C_2
Substituting the given values:
C_eq = 5.22 µF + 5.22 µF
Calculating this:
C_eq = 10.44 µF
Therefore, the equivalent capacitance between points a and b in Network 2 is 10.44 µF.
To summarize:
- Network 1: Equivalent capacitance = 2.61 µF
- Network 2: Equivalent capacitance = 10.44 µF