Physics, asked by GoLUKadora, 2 months ago

A generator having emf 100 volts is connected in series with 10 ohm resistor and an inductor of
2 Henry. If the switch is closed at a time t = 0, then the current at time t>O is
HOC​

Answers

Answered by Mithalesh1602398
0

Answer:

The current at time t > 0 is: i(t) = e^(-5t) * 50. This is the final solution for the current in the circuit.

Explanation:

The behavior of the circuit can be modeled by the differential equation:

L di/dt + R i = ε

where i is the current flowing through the circuit, L is the inductance, R is the resistance, ε is the emf, and t is time.

Substituting the given values, we get:

2 di/dt + 10 i = 100

To solve this differential equation, we can first find the homogeneous

2 di/dt + 10 i = 0

di/dt + 5 i = 0

This is a first-order linear homogeneous differential equation, which can be solved by separation of variables:

di/i = -5 dt

Integrating both sides, we get:

ln|i| = -5 t + C1

where C1 is the constant of integration. Exponentiating both sides, we get:

|i| = e^(-5t+C1) = e^(-5t) * e^(C1)

where e^(C1) is a positive constant.

Next, we need to find a particular solution to the non-homogeneous equation:

2 di/dt + 10 i = 100

A particular solution can be a constant, since the right-hand side of the equation is also a constant. Let i_p be this constant. Substituting this into the equation, we get:

2 di_p/dt + 10 i_p = 100

Solving for i_p, we get:

i_p = 50

Therefore, the general solution to the non-homogeneous equation is:

i = |i| * i_p = e^(-5t) * e^(C1) * 50

At t = 0, the switch is closed and the current is initially zero. Therefore, we can find the value of C1 as follows:

i(0) = e^(C1) * 50 = 0

Therefore, e^(C1) = 0, which is not possible for a positive constant. Hence, the initial current is zero.

Therefore, the current at time t > 0 is:

i(t) = e^(-5t) * 50

This is the final solution for the current in the circuit.The behavior of the circuit can be modeled by the differential equation:

L di/dt + R i = ε

where i is the current flowing through the circuit, L is the inductance, R is the resistance, ε is the emf, and t is time.

Substituting the given values, we get:

2 di/dt + 10 i = 100

To solve this differential equation, we can first find the homogeneous solution:

2 di/dt + 10 i = 0

di/dt + 5 i = 0

This is a first-order linear homogeneous differential equation, which can be solved by separation of variables:

di/i = -5 dt

Integrating both sides, we get:

ln|i| = -5 t + C1

where C1 is the constant of integration. Exponentiating both sides, we get:

|i| = e^(-5t+C1) = e^(-5t) * e^(C1)

where e^(C1) is a positive constant.

Next, we need to find a particular solution to the non-homogeneous equation:

2 di/dt + 10 i = 100

A particular solution can be a constant, since the right-hand side of the equation is also a constant. Let i_p be this constant. Substituting this into the equation, we get:

2 di_p/dt + 10 i_p = 100

Solving for i_p, we get:

i_p = 50

Therefore, the general solution to the non-homogeneous equation is:

i = |i| * i_p = e^(-5t) * e^(C1) * 50

At t = 0, the switch is closed and the current is initially zero. Therefore, we can find the value of C1 as follows:

i(0) = e^(C1) * 50 = 0

Therefore, e^(C1) = 0, which is not possible for a positive constant. Hence, the initial current is zero.

Therefore, the current at time t > 0 is:

i(t) = e^(-5t) * 50

This is the final solution for the current in the circuit.

To learn more about similar questions visit:

https://brainly.in/question/21171776?referrer=searchResults

https://brainly.in/question/21171776?referrer=searchResults

#SPJ1

Similar questions