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Difficulty: EasyCapacitors

An automotive audio technician installs two filter capacitors connected in parallel across a power line to suppress voltage ripple. If the individual capacitor values are 47 μF47\text{ }\mu\text{F} and 33 μF33\text{ }\mu\text{F}, what is the total equivalent capacitance of this parallel combination in microfarads (μF\mu\text{F})?

Answer: 80 \mu\text{F}

Answer

80 microfarads (80 μF80\text{ }\mu\text{F})
When capacitors are wired in parallel, the total plate area increases, allowing more charge to be stored at a given voltage. Therefore, the equivalent capacitance is the simple arithmetic sum of the individual capacitances: 47 μF+33 μF=80 μF47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}.

Step-by-Step Solution

1
Identify the network connection type and select the appropriate rule.
Capacitors are connected in parallel.
Connecting capacitors in parallel increases the total effective plate area, so their capacitances add directly.
2
Calculate the total capacitance using the sum formula Ctotal=C1+C2C_{\text{total}} = C_1 + C_2.
Ctotal=47 μF+33 μF=80 μFC_{\text{total}} = 47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}
Direct addition of the individual capacitance ratings yields the total circuit capacitance.

Key Concept

Parallel Capacitance Calculation
Estimated Time:45s
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Topics

Capacitors

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