
A capacitor is used to store \[24{\text{ watt hour}}\] of energy at \[1200{\text{ volt}}\]. What should be the capacitance of the capacitor ?
A. \[120{\text{ }}mF\]
B. \[120{\text{ }}\mu F\]
C. \[24{\text{ }}F\]
D. \[24{\text{ }}mF\]
Answer
463.2k+ views
Hint: The capacitor, like a miniature rechargeable battery, has the ability or "capacity" to store energy in the form of an electrical charge, producing a potential difference (Static Voltage) across its plates.
The Farad (abbreviated to $F$ ) is the unit of capacitance and is named after the British physicist Michael Faraday. Capacitance is the electrical property of a capacitor and is the measure of a capacitor's ability to store an electrical charge onto its two plates.
Formula Used:
The capacitance, \[C = \dfrac{Q}{V}\]
Where, $Q$ is the charge and $V$ is the potential difference.
The energy stored in capacitor is given by the formula,
\[E = \dfrac{1}{2}C{V^2}\]
Complete step by step answer:
The capacitance of a capacitor is defined as its ability to store an electric charge per unit of voltage across its plates. The formula \[C = \dfrac{Q}{V}\] is used to calculate capacitance by dividing electric charge by voltage. Farad is the name of the unit.For a capacitor,
The electrical energy stored,
\[E = \dfrac{1}{2}C{V^2}\]
Given,
\[E = 24{\text{watt - hour}}\]
$V = 1200{\text{ volt}}$
The unit of energy here is joule, so we have to convert energy form watt hour to joule.and, as
\[1{\text{watt - hour = 60}} \times {\text{60 = 3600\,joules}}\]
\[ \Rightarrow 24{\text{watt - hour = }}24 \times 60 \times 60{\text{ joules}}\]
So in the equation $E = \dfrac{1}{2}C{V^2}$ putting all the values we get,
\[ \Rightarrow 24 \times 60 \times 60 = \dfrac{1}{2} \times C \times 1200 \times 1200\]
\[ \Rightarrow C = 24 \times 2400\]
\[ \Rightarrow C = 12 \times {10^{ - 2}}F\]
\[ \therefore C\]\[ = 120 \times {10^{ - 3}}F\]
The capacitance will be \[120\,mF\].
Hence, the correct answer is option A.
Note: The capacitor charges up to its supply voltage when used in a direct current or DC circuit, but it prevents current flow since the dielectric of a capacitor is non-conductive and acts as an insulator. When a capacitor is linked to an alternating current (AC) circuit, however, the electricity seems to flow right through it with little or no resistance. Capacitors can also be employed to alter an audio circuit's frequency response or to connect separate amplifier stages that need to be protected from DC transmission.
The Farad (abbreviated to $F$ ) is the unit of capacitance and is named after the British physicist Michael Faraday. Capacitance is the electrical property of a capacitor and is the measure of a capacitor's ability to store an electrical charge onto its two plates.
Formula Used:
The capacitance, \[C = \dfrac{Q}{V}\]
Where, $Q$ is the charge and $V$ is the potential difference.
The energy stored in capacitor is given by the formula,
\[E = \dfrac{1}{2}C{V^2}\]
Complete step by step answer:
The capacitance of a capacitor is defined as its ability to store an electric charge per unit of voltage across its plates. The formula \[C = \dfrac{Q}{V}\] is used to calculate capacitance by dividing electric charge by voltage. Farad is the name of the unit.For a capacitor,
The electrical energy stored,
\[E = \dfrac{1}{2}C{V^2}\]
Given,
\[E = 24{\text{watt - hour}}\]
$V = 1200{\text{ volt}}$
The unit of energy here is joule, so we have to convert energy form watt hour to joule.and, as
\[1{\text{watt - hour = 60}} \times {\text{60 = 3600\,joules}}\]
\[ \Rightarrow 24{\text{watt - hour = }}24 \times 60 \times 60{\text{ joules}}\]
So in the equation $E = \dfrac{1}{2}C{V^2}$ putting all the values we get,
\[ \Rightarrow 24 \times 60 \times 60 = \dfrac{1}{2} \times C \times 1200 \times 1200\]
\[ \Rightarrow C = 24 \times 2400\]
\[ \Rightarrow C = 12 \times {10^{ - 2}}F\]
\[ \therefore C\]\[ = 120 \times {10^{ - 3}}F\]
The capacitance will be \[120\,mF\].
Hence, the correct answer is option A.
Note: The capacitor charges up to its supply voltage when used in a direct current or DC circuit, but it prevents current flow since the dielectric of a capacitor is non-conductive and acts as an insulator. When a capacitor is linked to an alternating current (AC) circuit, however, the electricity seems to flow right through it with little or no resistance. Capacitors can also be employed to alter an audio circuit's frequency response or to connect separate amplifier stages that need to be protected from DC transmission.
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