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A capacitor of capacitance 5μF is charged to a potential difference of 20V. After that, it is connected across an inductor of inductance 0.5 mH. What is the current flowing in the circuit at a time when the potential difference across the capacitor is 10 V?
- a)1 A
- b)2 A
- c)5 A
- d)0.5 A
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A capacitor of capacitance 5μF is charged to a potential difference...
Microfarads is connected in series with a resistor of resistance 10 ohms. If a voltage of 20 volts is applied across the circuit, what is the time constant of the circuit?
The time constant of an RC circuit is given by the formula:
τ = RC
where τ is the time constant in seconds, R is the resistance in ohms, and C is the capacitance in farads.
In this case, the capacitance is given as 5 microfarads, which is equivalent to 0.000005 farads. The resistance is given as 10 ohms. Therefore, the time constant is:
τ = RC = 10 ohms x 0.000005 farads = 0.00005 seconds
So the time constant of the circuit is 0.00005 seconds.
The time constant of an RC circuit is given by the formula:
τ = RC
where τ is the time constant in seconds, R is the resistance in ohms, and C is the capacitance in farads.
In this case, the capacitance is given as 5 microfarads, which is equivalent to 0.000005 farads. The resistance is given as 10 ohms. Therefore, the time constant is:
τ = RC = 10 ohms x 0.000005 farads = 0.00005 seconds
So the time constant of the circuit is 0.00005 seconds.
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A capacitor of capacitance 5μF is charged to a potential difference...
Given: V2 = 10 V; V1 = 20 V; C = 5 μF; Inductance (I) = 0.5 mH
Initial charge on the capacitor (q1) = C × V1 = 5 × 10-6 × 20 ……………….1
q1 = 10-4 C …………..B
The instantaneous charge on the capacitor as the capacitor discharges through the inductor ⇒ q2
q2 = q1cos (ωt) ⇒ q2/q2 = cos (ωt) ………………..A
Also, q2 = C × V2 = 5 × 10-6 × 10 …………………….2
q2 = 0.5 × 10-4 C
From 1 and 2 ⇒ q2/q2 = V2V1 ⇒ q2/q2 = 0.5 = 1/2
From equation A, we can equate as follows ⇒ cos (ωt) = 12
ωt = π/2 rad ………………..3
For an LC circuit ⇒
ω=20000rad/s …………………….4
The current through the circuit is given as:
Current (I)=-dq/dt
Charge decreases with respect to time, so, dq/dt obtained will be negative and this is why we add a negative sign to make a current
Initial charge on the capacitor (q1) = C × V1 = 5 × 10-6 × 20 ……………….1
q1 = 10-4 C …………..B
The instantaneous charge on the capacitor as the capacitor discharges through the inductor ⇒ q2
q2 = q1cos (ωt) ⇒ q2/q2 = cos (ωt) ………………..A
Also, q2 = C × V2 = 5 × 10-6 × 10 …………………….2
q2 = 0.5 × 10-4 C
From 1 and 2 ⇒ q2/q2 = V2V1 ⇒ q2/q2 = 0.5 = 1/2
From equation A, we can equate as follows ⇒ cos (ωt) = 12
ωt = π/2 rad ………………..3
For an LC circuit ⇒
ω=20000rad/s …………………….4
The current through the circuit is given as:
Current (I)=-dq/dt
Charge decreases with respect to time, so, dq/dt obtained will be negative and this is why we add a negative sign to make a current
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A capacitor of capacitance 5μF is charged to a potential difference of 20V. After that, it is connected across an inductor of inductance 0.5 mH. What is the current flowing in the circuit at a time when the potential difference across the capacitor is 10 V?a)1 Ab)2 Ac)5 Ad)0.5 ACorrect answer is option 'B'. Can you explain this answer?
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