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A coil of inductance 40 henry is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
- a)40 seconds
- b)20 seconds
- c)8 seconds
- d)5 seconds
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A coil of inductance 40 henry is connected in series with a resistance...
Time constant in in RL circuit is t = L/R
L = 40 henry
R = 8 ohm
t = L/R = 40/8 = 5 second
It is the time in which the current in RL circuit grows to 63.2 % of its peak value.
Most Upvoted Answer
A coil of inductance 40 henry is connected in series with a resistance...
Given:
- Inductance (L) = 40 H
- Resistance (R) = 8 Ω
- Voltage (V) = 2 V
To find: Time constant of the circuit
Solution:
The time constant (τ) of an RL circuit is given by the formula:
τ = L/R
where L is the inductance and R is the resistance.
Substituting the given values, we have:
τ = 40 H / 8 Ω
Simplifying the expression:
τ = 5 seconds
Therefore, the time constant of the circuit is 5 seconds.
Explanation:
An RL circuit consists of a resistor (R) and an inductor (L) connected in series. When a voltage is applied across the circuit, a current starts flowing through the circuit. However, due to the presence of inductance, the current does not immediately reach its maximum value.
The time constant of an RL circuit represents the time it takes for the current in the circuit to reach approximately 63.2% (1 - 1/e) of its maximum value. It is determined by the ratio of inductance to resistance (L/R).
In this case, the inductance is given as 40 H and the resistance is given as 8 Ω. By dividing the inductance by the resistance, we can calculate the time constant.
τ = L/R = 40 H / 8 Ω = 5 seconds
Hence, the time constant of the circuit is 5 seconds. This means that it takes approximately 5 seconds for the current to reach 63.2% of its maximum value in the circuit when a voltage of 2 V is applied.
- Inductance (L) = 40 H
- Resistance (R) = 8 Ω
- Voltage (V) = 2 V
To find: Time constant of the circuit
Solution:
The time constant (τ) of an RL circuit is given by the formula:
τ = L/R
where L is the inductance and R is the resistance.
Substituting the given values, we have:
τ = 40 H / 8 Ω
Simplifying the expression:
τ = 5 seconds
Therefore, the time constant of the circuit is 5 seconds.
Explanation:
An RL circuit consists of a resistor (R) and an inductor (L) connected in series. When a voltage is applied across the circuit, a current starts flowing through the circuit. However, due to the presence of inductance, the current does not immediately reach its maximum value.
The time constant of an RL circuit represents the time it takes for the current in the circuit to reach approximately 63.2% (1 - 1/e) of its maximum value. It is determined by the ratio of inductance to resistance (L/R).
In this case, the inductance is given as 40 H and the resistance is given as 8 Ω. By dividing the inductance by the resistance, we can calculate the time constant.
τ = L/R = 40 H / 8 Ω = 5 seconds
Hence, the time constant of the circuit is 5 seconds. This means that it takes approximately 5 seconds for the current to reach 63.2% of its maximum value in the circuit when a voltage of 2 V is applied.
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A coil of inductance 40 henry is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit isa)40 secondsb)20 secondsc)8 secondsd)5 secondsCorrect answer is option 'D'. Can you explain this answer?
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