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An ideal capacitor is charged to a voltage VO and connected at t=0 across an ideal inductor L. If ω = 1 / √LC, the voltage across the capacitor at time t > 0 is ________
- a)VO
- b)VO cos(ωt)
- c)VO sin(ωt)
- d)VO e-ωtcos(ωt)
Correct answer is option 'B'. Can you explain this answer?
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An ideal capacitor is charged to a voltage VOand connected at t=0 acro...
The voltage across the capacitor is VO, then the current flowing through the inductor when it is connected at t=0 will be zero. This is because at t=0, the voltage across the inductor is zero (since the capacitor is fully charged and there is no voltage drop across the inductor yet), and according to Ohm's Law (V = L * di/dt), the current flowing through the inductor will be zero.
However, as time progresses, the current through the inductor will start to build up. This is because the voltage across the capacitor will start to decrease, causing a voltage drop across the inductor. As the voltage across the capacitor decreases, the rate of change of current through the inductor (di/dt) increases, leading to an increasing current.
The rate at which the current builds up in the inductor is determined by the time constant of the circuit, which is given by the product of the inductance (L) and the resistance (R) of the circuit. In an ideal circuit with no resistance, the current will increase indefinitely. In a real circuit with resistance, the current will eventually reach a maximum value determined by the resistance.
In summary, when an ideal capacitor charged to a voltage VO is connected across an ideal inductor at t=0, the current through the inductor is initially zero, but will start to build up over time. The rate at which the current builds up depends on the time constant of the circuit.
However, as time progresses, the current through the inductor will start to build up. This is because the voltage across the capacitor will start to decrease, causing a voltage drop across the inductor. As the voltage across the capacitor decreases, the rate of change of current through the inductor (di/dt) increases, leading to an increasing current.
The rate at which the current builds up in the inductor is determined by the time constant of the circuit, which is given by the product of the inductance (L) and the resistance (R) of the circuit. In an ideal circuit with no resistance, the current will increase indefinitely. In a real circuit with resistance, the current will eventually reach a maximum value determined by the resistance.
In summary, when an ideal capacitor charged to a voltage VO is connected across an ideal inductor at t=0, the current through the inductor is initially zero, but will start to build up over time. The rate at which the current builds up depends on the time constant of the circuit.
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An ideal capacitor is charged to a voltage VOand connected at t=0 acro...
Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.
Voltage across capacitor will be decreasing from VO and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t > 0 is VO cos(ωt).
Voltage across capacitor will be decreasing from VO and periodic and is not decaying since both L and C is ideal.
∴ Voltage across the capacitor at time t > 0 is VO cos(ωt).
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An ideal capacitor is charged to a voltage VOand connected at t=0 across an ideal inductor L. If ω =1 / √LC, the voltage across the capacitor at time t > 0 is ________a)VOb)VOcos(ωt)c)VOsin(ωt)d)VOe-ωtcos(ωt)Correct answer is option 'B'. Can you explain this answer?
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