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
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.
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