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A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F?
Verified Answer
A capacitor of capacitance C is charged with the help of a 200V batter...
Option (B) seems to be right.
Electrical potential Energy stored in the capacitor = 1/2 C V^2
= U = 1/2 * C * 200^2 = 20,000 C Joules
Energy dissipated through the resistor = heat absorbed by the block
H = 2.5 * 10^2 J/kg/degreeK * 0.1 Kg * 0.4degreeK
= 10 J
U = H => C = 10/20,000 F = 500 * 10⁻^6 F
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Most Upvoted Answer
A capacitor of capacitance C is charged with the help of a 200V batter...
Given:
- Capacitance (C)
- Voltage of the battery (V = 200V)
- Resistance of the coil (R)
- Specific heat capacity of the block (c = 2.5 x 10^2 J/kg)
- Mass of the block (m = 0.1kg)
- Change in temperature of the block (ΔT = 0.4K)
To find:
The value of capacitance (C)
Formula:
The energy stored in a capacitor is given by the formula:
E = 0.5 * C * V^2
The energy transferred to the block is given by the formula:
E = mcΔT
The power dissipated by the coil is given by the formula:
P = I^2 * R
The current flowing through the coil is given by the formula:
I = V / R
Solution:
Step 1: Calculate the energy stored in the capacitor during charging.
Given that the voltage of the battery is 200V and the energy stored in the capacitor is given by E = 0.5 * C * V^2, we can calculate the energy as:
E1 = 0.5 * C * (200)^2
Step 2: Calculate the energy transferred to the block during discharge.
Given that the mass of the block is 0.1kg, the specific heat capacity is 2.5 x 10^2 J/kg, and the change in temperature is 0.4K, we can calculate the energy transferred as:
E2 = mcΔT
Step 3: Calculate the power dissipated by the coil during discharge.
Given that the resistance of the coil is R, the current flowing through the coil is I = V / R, and the power dissipated is P = I^2 * R, we can calculate the power as:
P = (V / R)^2 * R
Step 4: Equate the energy transferred to the block with the energy stored in the capacitor.
Since the energy transferred to the block during discharge is equal to the energy stored in the capacitor during charging, we can equate E1 and E2 and solve for C:
0.5 * C * (200)^2 = mcΔT
Step 5: Simplify and solve for C.
C = (2mcΔT) / (V^2)
Now, substitute the given values into the equation to find the value of C.
C = (2 * 0.1kg * 2.5 x 10^2 J/kg * 0.4K) / (200V)^2
C = 50 micro F
Answer:
The value of capacitance (C) is 50 micro F. (Option d)
- Capacitance (C)
- Voltage of the battery (V = 200V)
- Resistance of the coil (R)
- Specific heat capacity of the block (c = 2.5 x 10^2 J/kg)
- Mass of the block (m = 0.1kg)
- Change in temperature of the block (ΔT = 0.4K)
To find:
The value of capacitance (C)
Formula:
The energy stored in a capacitor is given by the formula:
E = 0.5 * C * V^2
The energy transferred to the block is given by the formula:
E = mcΔT
The power dissipated by the coil is given by the formula:
P = I^2 * R
The current flowing through the coil is given by the formula:
I = V / R
Solution:
Step 1: Calculate the energy stored in the capacitor during charging.
Given that the voltage of the battery is 200V and the energy stored in the capacitor is given by E = 0.5 * C * V^2, we can calculate the energy as:
E1 = 0.5 * C * (200)^2
Step 2: Calculate the energy transferred to the block during discharge.
Given that the mass of the block is 0.1kg, the specific heat capacity is 2.5 x 10^2 J/kg, and the change in temperature is 0.4K, we can calculate the energy transferred as:
E2 = mcΔT
Step 3: Calculate the power dissipated by the coil during discharge.
Given that the resistance of the coil is R, the current flowing through the coil is I = V / R, and the power dissipated is P = I^2 * R, we can calculate the power as:
P = (V / R)^2 * R
Step 4: Equate the energy transferred to the block with the energy stored in the capacitor.
Since the energy transferred to the block during discharge is equal to the energy stored in the capacitor during charging, we can equate E1 and E2 and solve for C:
0.5 * C * (200)^2 = mcΔT
Step 5: Simplify and solve for C.
C = (2mcΔT) / (V^2)
Now, substitute the given values into the equation to find the value of C.
C = (2 * 0.1kg * 2.5 x 10^2 J/kg * 0.4K) / (200V)^2
C = 50 micro F
Answer:
The value of capacitance (C) is 50 micro F. (Option d)
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A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F?
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A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F?.
A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A capacitor of capacitance C is charged with the help of a 200V battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 2.5 x 10^2 J/kg and mass 0.1kg. If the temperature of the block roses by 0.4K, the value of C is a) 500F b) 500 micro F c) 50 F d) 50 micro F?.
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