To determine the equivalent capacitance of the given series-parallel combination of capacitors, let's analyze the circuit step-by-step.

Step-by-Step Solution

1. Identify Capacitors in Series and Parallel:

   • From the diagram:
     • $C_1 = 2 \, \mu\text{F}$
     • $C_2 = 3 \, \mu\text{F}$
     • $C_3 = 4 \, \mu\text{F}$
     • $C_4 = 5 \, \mu\text{F}$

2. Combine Capacitors $C_2$ and $C_3$ in Series:

   • When capacitors are in series, the equivalent capacitance $C_{23}$ is given by: $\frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3}$ Substituting values: $\frac{1}{C_{23}} = \frac{1}{3 \, \mu\text{F}} + \frac{1}{4 \, \mu\text{F}}$ $\frac{1}{C_{23}} = \frac{4 + 3}{12} = \frac{7}{12}$ Therefore: $C_{23} = \frac{12}{7} \approx 1.71 \, \mu\text{F}$

3. Combine $C_{23}$ with $C_4$ in Parallel:

   • When capacitors are in parallel, the equivalent capacitance $C_{234}$ is: $C_{234} = C_{23} + C_4$ Substituting values: $C_{234} = 1.71 \, \mu\text{F} + 5 \, \mu\text{F} = 6.71 \, \mu\text{F}$

4. Combine $C_{234}$ with $C_1$ in Series:

   • Now, we combine $C_{234}$ and $C_1$ in series: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_{234}}$ Substituting values: $\frac{1}{C_{\text{eq}}} = \frac{1}{2 \, \mu\text{F}} + \frac{1}{6.71 \, \mu\text{F}}$ $\frac{1}{C_{\text{eq}}} = \frac{6.71 + 2}{13.42} = \frac{8.71}{13.42}$ Therefore: $C_{\text{eq}} \approx 1.18 \, \mu\text{F}$

Answer

The resulting capacitance of the circuit is: $C_{\text{eq}} = 1.18 \, \mu\text{F}$

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Related Questions:

1. What would happen to the equivalent capacitance if $C_1$ were increased?
2. How do you calculate equivalent capacitance for purely parallel combinations?
3. What changes if $C_2$ and $C_3$ are arranged in parallel instead of series?
4. Can you explain why capacitors in series result in a smaller equivalent capacitance?
5. How does adding a capacitor in parallel affect the total capacitance?

Tip:

For series capacitors, the equivalent capacitance is always less than the smallest individual capacitor in the series.