When five capacitors with equal capacitances are connected in series, the equivalent capacitance of the combination is 8.31 mF. The capacitors are then reconnected so that a parallel combination of two capacitors is connected in series with a parallel combination of three capacitors.
The Correct Answer and Explanation is :
To solve the problem, we first need to determine the capacitance of each capacitor when connected in series. Then, we’ll analyze the new combination of capacitors.
Step 1: Equivalent Capacitance in Series
For capacitors in series, the equivalent capacitance ((C_{\text{eq}})) is given by:
[
\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4} + \frac{1}{C_5}
]
Since all five capacitors have equal capacitances ((C)):
[
\frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{5}{C}
]
Given (C_{\text{eq}} = 8.31 \, \text{mF}), we solve for (C):
[
C = 5 \times C_{\text{eq}} = 5 \times 8.31 \, \text{mF} = 41.55 \, \text{mF}
]
Each capacitor has a capacitance of (C = 41.55 \, \text{mF}).
Step 2: New Combination
The capacitors are reconnected with a parallel combination of two capacitors ((C_{\text{parallel1}})) in series with a parallel combination of three capacitors ((C_{\text{parallel2}})).
Capacitance of Parallel Groups:
For parallel capacitors, the equivalent capacitance is the sum:
[
C_{\text{parallel1}} = C + C = 2C = 2 \times 41.55 = 83.1 \, \text{mF}
]
[
C_{\text{parallel2}} = C + C + C = 3C = 3 \times 41.55 = 124.65 \, \text{mF}
]
Capacitance in Series:
For the series connection of (C_{\text{parallel1}}) and (C_{\text{parallel2}}):
[
\frac{1}{C_{\text{eq-new}}} = \frac{1}{C_{\text{parallel1}}} + \frac{1}{C_{\text{parallel2}}}
]
Substitute the values:
[
\frac{1}{C_{\text{eq-new}}} = \frac{1}{83.1} + \frac{1}{124.65}
]
Calculate:
[
\frac{1}{C_{\text{eq-new}}} = 0.01203 + 0.00802 = 0.02005
]
[
C_{\text{eq-new}} = \frac{1}{0.02005} = 49.9 \, \text{mF}
]
Final Answer:
The equivalent capacitance of the new combination is 49.9 mF.
Explanation:
Capacitors in series result in a lower equivalent capacitance than any single capacitor, as the total charge has to be shared. When capacitors are in parallel, the total capacitance is the sum, as the effective plate area increases. By strategically combining capacitors in series and parallel, the new configuration balances these principles, yielding a higher equivalent capacitance compared to the original series arrangement.