Calculate the capacitance of a capacitor given a charge of 24 microcoulombs at a voltage of 12 volts.
The Correct Answer and Explanation is:
To calculate the capacitance of a capacitor, we can use the formula for capacitance:C=QVC = \frac{Q}{V}C=VQ
Where:
- CCC is the capacitance in farads (F),
- QQQ is the charge on the capacitor in coulombs (C),
- VVV is the voltage across the capacitor in volts (V).
Given:
- Q=24 μC=24×10−6 CQ = 24 \, \mu C = 24 \times 10^{-6} \, CQ=24μC=24×10−6C,
- V=12 VV = 12 \, VV=12V.
Now, substitute the values into the formula:C=24×10−6 C12 VC = \frac{24 \times 10^{-6} \, C}{12 \, V}C=12V24×10−6CC=2×10−6 FC = 2 \times 10^{-6} \, FC=2×10−6FC=2 μFC = 2 \, \mu FC=2μF
Conclusion:
The capacitance of the capacitor is 2 μF2 \, \mu F2μF (microfarads).
Explanation:
Capacitance is the ability of a capacitor to store charge. The formula C=QVC = \frac{Q}{V}C=VQ shows that capacitance is directly proportional to the charge stored and inversely proportional to the voltage applied.
A larger charge or a lower voltage would result in a higher capacitance. Conversely, a smaller charge or a higher voltage would result in a lower capacitance. This relationship is fundamental in capacitor design, where capacitors are used to store energy or smooth voltage fluctuations in circuits. The unit of capacitance, the farad, is often impractically large for most everyday applications, so microfarads (μF\mu FμF) and nanofarads (nFnFnF) are more commonly used for practical capacitors.
