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The terminal voltage of a 2- H inductor is v = 10(1-t) v . Find the current flowing through it at t= 4s and the energy atored in it at t= 4 S. Assume I(0) =2A. Please provide the detailed solution?
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The terminal voltage of a 2- H inductor is v = 10(1-t) v . Find the cu...
**Given:**
- Terminal voltage of a 2 H inductor: v = 10(1 - t) V
- Time: t = 4 s
- Initial current: I(0) = 2 A
**To find:**
1. Current flowing through the inductor at t = 4 s
2. Energy stored in the inductor at t = 4 s
**Solution:**
1. Current flowing through the inductor at t = 4 s:
- The voltage across an inductor is given by the equation V = L(di/dt), where V is the voltage, L is the inductance, and di/dt is the rate of change of current.
- Rearranging the equation, we have di = V/L * dt.
- Integrating both sides of the equation, we get ∫di = ∫V/L * dt.
- Let's integrate the equation within the given limits (from 0 to t) to find the current flowing through the inductor at t = 4 s.
- ∫(2 to i)di = ∫(0 to 4)10(1 - t)/2 dt.
- Integrating, we get (i - 2) = [10t - 5t^2/2] (0 to 4).
- Substituting the values, we get i - 2 = [10(4) - 5(4^2)/2] - [10(0) - 5(0^2)/2].
- Simplifying, we have i - 2 = [40 - 5(8)] - [0 - 0].
- i - 2 = 40 - 40.
- i - 2 = 0.
- i = 2 A.
2. Energy stored in the inductor at t = 4 s:
- The energy stored in an inductor is given by the equation W = (1/2)Li^2, where W is the energy, L is the inductance, and i is the current flowing through the inductor.
- Substituting the values, we have W = (1/2) * 2 * (2^2).
- Simplifying, we get W = 2 J.
**Conclusion:**
- The current flowing through the inductor at t = 4 s is 2 A.
- The energy stored in the inductor at t = 4 s is 2 J.
- Terminal voltage of a 2 H inductor: v = 10(1 - t) V
- Time: t = 4 s
- Initial current: I(0) = 2 A
**To find:**
1. Current flowing through the inductor at t = 4 s
2. Energy stored in the inductor at t = 4 s
**Solution:**
1. Current flowing through the inductor at t = 4 s:
- The voltage across an inductor is given by the equation V = L(di/dt), where V is the voltage, L is the inductance, and di/dt is the rate of change of current.
- Rearranging the equation, we have di = V/L * dt.
- Integrating both sides of the equation, we get ∫di = ∫V/L * dt.
- Let's integrate the equation within the given limits (from 0 to t) to find the current flowing through the inductor at t = 4 s.
- ∫(2 to i)di = ∫(0 to 4)10(1 - t)/2 dt.
- Integrating, we get (i - 2) = [10t - 5t^2/2] (0 to 4).
- Substituting the values, we get i - 2 = [10(4) - 5(4^2)/2] - [10(0) - 5(0^2)/2].
- Simplifying, we have i - 2 = [40 - 5(8)] - [0 - 0].
- i - 2 = 40 - 40.
- i - 2 = 0.
- i = 2 A.
2. Energy stored in the inductor at t = 4 s:
- The energy stored in an inductor is given by the equation W = (1/2)Li^2, where W is the energy, L is the inductance, and i is the current flowing through the inductor.
- Substituting the values, we have W = (1/2) * 2 * (2^2).
- Simplifying, we get W = 2 J.
**Conclusion:**
- The current flowing through the inductor at t = 4 s is 2 A.
- The energy stored in the inductor at t = 4 s is 2 J.
Community Answer
The terminal voltage of a 2- H inductor is v = 10(1-t) v . Find the cu...
I= -18 A & 324 J energy
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The terminal voltage of a 2- H inductor is v = 10(1-t) v . Find the current flowing through it at t= 4s and the energy atored in it at t= 4 S. Assume I(0) =2A. Please provide the detailed solution?
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