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Two capacitors A and B of capacitance C and 2C are connected in series across a potential difference V. Find the ratio of the energy stored in capacitor A to capacitor B.
- a)1 : 2
- b)2 : 1
- c)1 : 4
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
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Two capacitors A and B of capacitance C and 2C are connected in series...
CONCEPT:
Capacitor:
The capacitor is a device in which electrical energy can be stored.
- In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite signs and separated by an insulating medium.
- The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.
Capacitance:
Capacitance:
- The property to store energy in form of charge is called capacitance.
- The unit of capacitance is Farad.
- The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,
⇒ Q ∝ V
⇒ Q = CV
Where C = capacitance
Energy stored in the capacitor:
- If a capacitor of capacitance C is charged by a potential difference V, then the energy stored in the capacitor is given as,
CALCULATION:
Given QA = QB = Q, CA = C, and CB = 2C
- We know that when capacitors are connected in series, the magnitude of charge Q on each capacitor is the same.
- We know that the energy stored in the capacitor is given as,
By equation 1 for the capacitor A,
By equation 1 for the capacitor B,
By equation 2 and equation 3,
Hence option 2 is correct.
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Community Answer
Two capacitors A and B of capacitance C and 2C are connected in series...
Explanation:
Given:
- Capacitance of capacitor A = C
- Capacitance of capacitor B = 2C
- Potential difference across capacitors = V
Energy stored in a capacitor:
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} CV^2 \]
Calculating energy stored in capacitor A:
\[ U_A = \frac{1}{2} CV^2 \]
Calculating energy stored in capacitor B:
\[ U_B = \frac{1}{2} (2C)V^2 = CV^2 \]
Ratio of energy stored in capacitor A to capacitor B:
\[ \frac{U_A}{U_B} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} \]
Therefore, the ratio of the energy stored in capacitor A to capacitor B is 1:2.
So, the correct answer is option 2 : 1.
Given:
- Capacitance of capacitor A = C
- Capacitance of capacitor B = 2C
- Potential difference across capacitors = V
Energy stored in a capacitor:
The energy stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} CV^2 \]
Calculating energy stored in capacitor A:
\[ U_A = \frac{1}{2} CV^2 \]
Calculating energy stored in capacitor B:
\[ U_B = \frac{1}{2} (2C)V^2 = CV^2 \]
Ratio of energy stored in capacitor A to capacitor B:
\[ \frac{U_A}{U_B} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} \]
Therefore, the ratio of the energy stored in capacitor A to capacitor B is 1:2.
So, the correct answer is option 2 : 1.
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Two capacitors A and B of capacitance C and 2C are connected in series across a potential difference V. Find the ratio of the energy stored in capacitor A to capacitor B.a)1 : 2b)2 : 1c)1 : 4d)None of theseCorrect answer is option 'B'. Can you explain this answer?
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