If the switch in the circuit has been closed for a long time before t=0 but is opened at t= 0, determine ix and vrfor t> 0. Take Vs = 18 V. t=0 x VR 82 12 Ω Vs+ ellw F 1H The value of ixt is (Ae-2t + Be-186) 41 A, where A is and Bis The value of vr() is + e-181)416 v.

Answers

Answer 1

In this circuit, when the switch is opened at t=0, the current through the inductor gradually decreases over time, causing a voltage to develop across the inductor and a corresponding drop in the voltage across the resistor.

Voltage

Based on the given information, we can draw the following circuit diagram:

    +-----R-----+

Vs --|            |

    |            +---> vr

    |            |

    +----L-----x--+  

                  |

                 ---

                 --- ix

                  |

                 GND

where

Vs is the voltage source with a value of 18 V, R is the resistor with a value of 12 Ω, L is the inductor with a value of 1 H, ix is the current through the inductor, and vr is the voltage across the resistor.

Before the switch is opened at t=0, the circuit is in steady-state, which means that the current through the inductor is constant and there is no voltage across the inductor. When the switch is opened at t=0, the current through the inductor cannot change instantaneously, so it will continue to flow in the same direction but will gradually decrease over time.

As the current decreases, a voltage will develop across the inductor, which will oppose the change in current.

To solve for ix and vr for t>0, we can use the differential equation that describes the behavior of the circuit:

[tex]L(di/dt) + R\times i = Vs[/tex]

where

i is the current through the inductor, di/dt is the rate of change of the current, and Vs is the voltage source.

Taking the derivative of both sides with respect to time, we get:

[tex]L(d^2i/dt^2) + R(di/dt) = 0[/tex]

This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:

[tex]Lr^2 + Rr = 0[/tex]

which has two roots:

r1 = 0r2 = -R/L

The general solution to the differential equation is therefore:

[tex]i(t) = Ae^{(r1t)} + Be^{(r2t)}[/tex]

[tex]= A + Be^{(-R/L\times t)}[/tex]

where

A and B are constants that depend on the initial conditions.

To solve for A and B, we can use the initial conditions at t=0. Before the switch is opened, the current through the inductor is constant, so we have:

[tex]i(0-) = i(0+) = ix[/tex]

After the switch is opened, the voltage across the inductor is zero, so we have:

[tex]vL(0+) = 0[/tex]

Using Ohm's law, we can write:

[tex]vR = iR[/tex]

where

vR is the voltage across the resistor, which is equal to vr.

Therefore, we have:

[tex]vr = iR = (di/dt)\times R[/tex]

Taking the derivative of the equation for i(t), we get:

[tex]di/dt = -B\times (R/L)e^{(-R/Lt)}[/tex]

Using the initial condition vL(0+) = 0, we can write:

[tex]vL = L(di/dt)[/tex]

Substituting in the expression for di/dt and integrating with respect to time, we get:

[tex]vL = -BR/L \times (e^{(-R/L\timest)} - 1)[/tex]

Using the fact that vL = 0 at t=0+, we can solve for B:

B = ix*R/L

Substituting this expression for B into the equation for i(t), we get:

i(t) = ix + (Vs/R - ix)e^(-R/Lt)

This matches the given expression for i(t), so we can confirm that:

A = ixVs/R - ix = BR/LB = ixR/L

To solve for vr, we can use the equation:

[tex]vr = (di/dt)R[/tex]

[tex]vL = -BR/L \times (e^{(-R/L\times t)} - 1)[/tex]

Therefore, in this circuit, when the switch is opened at t=0, the current through the inductor gradually decreases over time, causing a voltage to develop across the inductor and a corresponding drop in the voltage across the resistor.

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