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. A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later?
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. A direct voltage of 200 V is suddenly applied to a series L-R circui...
When a direct voltage is suddenly applied to a series L-R circuit, the voltage drop across the inductor will depend on the inductance and resistance of the circuit and the time elapsed since the voltage was applied.
To determine the voltage drop across the inductor at the instant of switching on and at 0.02 seconds later, we can use the following equation:
V = L(di/dt) + Ri
where V is the voltage drop across the inductor, L is the inductance of the circuit, di/dt is the rate of change of current, R is the resistance of the circuit, and i is the current in the circuit.
In this case, the circuit has an inductance of 0.2 H and a resistance of 20 ohms. The direct voltage applied to the circuit is 200 V.
At the instant of switching on, the current in the circuit is zero, so the voltage drop across the inductor is also zero.
At 0.02 seconds later, the current in the circuit will have increased, and the voltage drop across the inductor will be given by the equation above. To determine the voltage drop at this time, we need to know the current in the circuit.
The current in the circuit can be found using Ohm's law, which states that the current in a circuit is equal to the voltage divided by the resistance:
I = V/R
Plugging in the values given, we find that the current in the circuit is 200 V / 20 ohms = 10 A.
Substituting this value into the equation for the voltage drop across the inductor, we find that the voltage drop at 0.02 seconds later is:
V = (0.2 H)(10 A/s) + (20 ohms)(10 A) = 4 V + 200 V = 204 V
Therefore, the voltage drop across the inductor at the instant of switching on is 0 V, and the voltage drop at 0.02 seconds later is 204 V.
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. A direct voltage of 200 V is suddenly applied to a series L-R circui...
Solution:
Given:
- Direct voltage (V) = 200 V
- Resistance (R) = 20 Ω
- Inductance (L) = 0.2 H (given as 0 2 H)
To find:
- Voltage drop across the inductor at the instant of switching on and at 0.02 sec later
Formula:
The formula to calculate the voltage drop across an inductor is given by:
V = L(di/dt)
Where,
- V is the voltage drop across the inductor
- L is the inductance of the circuit
- di/dt is the rate of change of current with time
Calculation:
1. Voltage drop at the instant of switching on:
At the instant of switching on, the current through the inductor is zero, and hence the rate of change of current is also zero (di/dt = 0). Therefore, the voltage drop across the inductor at this instant is also zero.
2. Voltage drop at 0.02 sec later:
To calculate the voltage drop at 0.02 sec later, we need to find the rate of change of current (di/dt) at that instant.
We know that the voltage across the inductor is equal to the rate of change of current multiplied by the inductance (V = L(di/dt)). Rearranging the equation, we have:
di/dt = V/L
Substituting the given values, we have:
di/dt = 200 V / 0.2 H
di/dt = 1000 A/s
Now, we can calculate the voltage drop across the inductor at 0.02 sec later using the formula V = L(di/dt):
V = 0.2 H * 1000 A/s
V = 200 V
Therefore, the voltage drop across the inductor at 0.02 sec later is 200 V.
Summary:
- At the instant of switching on, the voltage drop across the inductor is zero.
- At 0.02 sec later, the voltage drop across the inductor is 200 V.
Given:
- Direct voltage (V) = 200 V
- Resistance (R) = 20 Ω
- Inductance (L) = 0.2 H (given as 0 2 H)
To find:
- Voltage drop across the inductor at the instant of switching on and at 0.02 sec later
Formula:
The formula to calculate the voltage drop across an inductor is given by:
V = L(di/dt)
Where,
- V is the voltage drop across the inductor
- L is the inductance of the circuit
- di/dt is the rate of change of current with time
Calculation:
1. Voltage drop at the instant of switching on:
At the instant of switching on, the current through the inductor is zero, and hence the rate of change of current is also zero (di/dt = 0). Therefore, the voltage drop across the inductor at this instant is also zero.
2. Voltage drop at 0.02 sec later:
To calculate the voltage drop at 0.02 sec later, we need to find the rate of change of current (di/dt) at that instant.
We know that the voltage across the inductor is equal to the rate of change of current multiplied by the inductance (V = L(di/dt)). Rearranging the equation, we have:
di/dt = V/L
Substituting the given values, we have:
di/dt = 200 V / 0.2 H
di/dt = 1000 A/s
Now, we can calculate the voltage drop across the inductor at 0.02 sec later using the formula V = L(di/dt):
V = 0.2 H * 1000 A/s
V = 200 V
Therefore, the voltage drop across the inductor at 0.02 sec later is 200 V.
Summary:
- At the instant of switching on, the voltage drop across the inductor is zero.
- At 0.02 sec later, the voltage drop across the inductor is 200 V.
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. A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about . A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later?.
. A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about . A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for . A direct voltage of 200 V is suddenly applied to a series L-R circuit having R=20 ohms, and inductance 0 2 H. Determine the voltage drop across the inductor at the instant of switching on and at 0.02 sec later?.
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