When five capacitors with equal capacitances are connected in series, the equivalent capacitance of the combination is 6.99 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. Determine the equivalent capacitance Cequiv of this combination in millifarads. Cequiv mF PC TOOLS X10
The Correct Answer and Explanation is:
Let’s first address the scenario where five capacitors, each with the same capacitance ( C ), are connected in series. The formula for the equivalent capacitance of capacitors in series is:
[
\frac{1}{C_{\text{equiv, series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}
]
Since the five capacitors have equal capacitances, the formula simplifies to:
[
\frac{1}{C_{\text{equiv, series}}} = \frac{5}{C}
]
Thus, the equivalent capacitance for five capacitors in series is:
[
C_{\text{equiv, series}} = \frac{C}{5}
]
We’re given that the equivalent capacitance for the five capacitors in series is 6.99 mF. Therefore:
[
\frac{C}{5} = 6.99 \, \text{mF}
]
Solving for ( C ), the capacitance of each individual capacitor:
[
C = 6.99 \, \text{mF} \times 5 = 34.95 \, \text{mF}
]
Now, the capacitors are reconnected in a different configuration: two capacitors are connected in parallel, and this parallel combination is connected in series with another parallel combination of three capacitors.
Step 1: Equivalent Capacitance for Two Capacitors in Parallel
The formula for capacitors in parallel is:
[
C_{\text{equiv, parallel}} = C_1 + C_2 + \cdots + C_n
]
For two capacitors, the equivalent capacitance is:
[
C_{\text{parallel, 2}} = C + C = 2C = 2 \times 34.95 \, \text{mF} = 69.9 \, \text{mF}
]
Step 2: Equivalent Capacitance for Three Capacitors in Parallel
For three capacitors, the equivalent capacitance is:
[
C_{\text{parallel, 3}} = C + C + C = 3C = 3 \times 34.95 \, \text{mF} = 104.85 \, \text{mF}
]
Step 3: Equivalent Capacitance of Two Parallel Combinations in Series
Finally, we need to find the equivalent capacitance of these two parallel combinations connected in series. The formula for two capacitors in series is:
[
\frac{1}{C_{\text{equiv, series}}} = \frac{1}{C_1} + \frac{1}{C_2}
]
Substituting ( C_1 = 69.9 \, \text{mF} ) and ( C_2 = 104.85 \, \text{mF} ):
[
\frac{1}{C_{\text{equiv, series}}} = \frac{1}{69.9} + \frac{1}{104.85}
]
Calculating the individual terms:
[
\frac{1}{69.9} \approx 0.0143 \quad \text{and} \quad \frac{1}{104.85} \approx 0.0095
]
Thus:
[
\frac{1}{C_{\text{equiv, series}}} = 0.0143 + 0.0095 = 0.0238
]
Finally, the equivalent capacitance is:
[
C_{\text{equiv, series}} = \frac{1}{0.0238} \approx 42.02 \, \text{mF}
]
Final Answer:
The equivalent capacitance of the combination is approximately 42.02 mF.