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Consider a 2F capacitor having an initial voltage of 2 volts at time t=0- . If the current flowing through the capacitor is 2δ(t) , then the energy (in joules) stored in the capacitor at time t=0+ will be _______. (in integer)
Correct answer is '9'. Can you explain this answer?
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Consider a 2F capacitor having an initial voltage of 2 volts at time ...
Given:
- Capacitance (C) = 2F
- Initial voltage (V) = 2V
- Current (i) = 2δ(t), where δ(t) is the Dirac delta function
To find:
- Energy stored in the capacitor at time t = 0
Explanation:
1. Energy stored in a capacitor:
The energy stored in a capacitor can be calculated using the formula:
E = 0.5 * C * V^2
Where:
- E is the energy stored in the capacitor
- C is the capacitance of the capacitor
- V is the voltage across the capacitor
2. Current through the capacitor:
The current through a capacitor is given by the formula:
i = C * dv/dt
Where:
- i is the current through the capacitor
- C is the capacitance of the capacitor
- dv/dt is the rate of change of voltage with respect to time
In this case, the current through the capacitor is given as 2δ(t). Since δ(t) is the Dirac delta function, it represents an impulse of infinite magnitude at t = 0.
3. Relationship between current and voltage:
To find the voltage across the capacitor, we need to integrate the current.
∫i dt = ∫C * dv/dt dt
Integrating both sides:
∫2δ(t) dt = C * ∫dv
2 * ∫δ(t) dt = C * v
2 * 1 = C * v
2 = 2F * v
v = 1V
4. Calculating energy:
Now that we have the voltage across the capacitor (v = 1V), we can calculate the energy stored in the capacitor using the formula mentioned earlier.
E = 0.5 * C * V^2
Substituting the values:
E = 0.5 * 2F * (1V)^2
E = 0.5 * 2J
E = 1J
Therefore, the energy stored in the capacitor at time t = 0 is 1 joule, which is not equal to 9 as mentioned in the correct answer.
- Capacitance (C) = 2F
- Initial voltage (V) = 2V
- Current (i) = 2δ(t), where δ(t) is the Dirac delta function
To find:
- Energy stored in the capacitor at time t = 0
Explanation:
1. Energy stored in a capacitor:
The energy stored in a capacitor can be calculated using the formula:
E = 0.5 * C * V^2
Where:
- E is the energy stored in the capacitor
- C is the capacitance of the capacitor
- V is the voltage across the capacitor
2. Current through the capacitor:
The current through a capacitor is given by the formula:
i = C * dv/dt
Where:
- i is the current through the capacitor
- C is the capacitance of the capacitor
- dv/dt is the rate of change of voltage with respect to time
In this case, the current through the capacitor is given as 2δ(t). Since δ(t) is the Dirac delta function, it represents an impulse of infinite magnitude at t = 0.
3. Relationship between current and voltage:
To find the voltage across the capacitor, we need to integrate the current.
∫i dt = ∫C * dv/dt dt
Integrating both sides:
∫2δ(t) dt = C * ∫dv
2 * ∫δ(t) dt = C * v
2 * 1 = C * v
2 = 2F * v
v = 1V
4. Calculating energy:
Now that we have the voltage across the capacitor (v = 1V), we can calculate the energy stored in the capacitor using the formula mentioned earlier.
E = 0.5 * C * V^2
Substituting the values:
E = 0.5 * 2F * (1V)^2
E = 0.5 * 2J
E = 1J
Therefore, the energy stored in the capacitor at time t = 0 is 1 joule, which is not equal to 9 as mentioned in the correct answer.
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Community Answer
Consider a 2F capacitor having an initial voltage of 2 volts at time ...
Let Vc (t) be the voltage across the 2F capacitor and let Ic (t) be the current through the capacitor.
We know that,
At t=0+ ,
Given:
C=2F
Vc(0-) =2V
And Ic (t)= 2δ(t)
Since, 
Vc(0+) =2+1 =3V
∴ Energy stored in the capacitor at t=0+ will be,
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Consider a 2F capacitor having an initial voltage of 2 volts at time t=0- . If the current flowing through the capacitor is 2δ(t) , then the energy (in joules) stored in the capacitor at time t=0+ will be _______. (in integer)Correct answer is '9'. Can you explain this answer?
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Consider a 2F capacitor having an initial voltage of 2 volts at time t=0- . If the current flowing through the capacitor is 2δ(t) , then the energy (in joules) stored in the capacitor at time t=0+ will be _______. (in integer)Correct answer is '9'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Consider a 2F capacitor having an initial voltage of 2 volts at time t=0- . If the current flowing through the capacitor is 2δ(t) , then the energy (in joules) stored in the capacitor at time t=0+ will be _______. (in integer)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a 2F capacitor having an initial voltage of 2 volts at time t=0- . If the current flowing through the capacitor is 2δ(t) , then the energy (in joules) stored in the capacitor at time t=0+ will be _______. (in integer)Correct answer is '9'. Can you explain this answer?.
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