The switch in the circuit has been in position “a” for a long time. At time t = 0, it moves instantaneously to position “b”. For t ≥ 0+, find: v0 (t) = 11. 80 1/1 points + 12. -40 1/1 points e( 13. -1000 1/1 points t) V i0 (t) = 14. 0
The Correct Answer and Explanation is:
It seems like you’re dealing with an RL circuit where the switch changes position at time t=0t = 0t=0, transitioning from position “a” to position “b”. We need to calculate the voltage and current as functions of time after the switch moves. This type of circuit is typically analyzed using differential equations for the circuit components.
Given:
- The initial voltage v0(t)v_0(t)v0(t) and initial current i0(t)i_0(t)i0(t).
- The voltage for t≥0t \geq 0t≥0 is expressed as v0(t)=11.80−40e−1000tv_0(t) = 11.80 – 40 e^{-1000t}v0(t)=11.80−40e−1000t volts.
- The current for t≥0t \geq 0t≥0 is i0(t)=14.0i_0(t) = 14.0i0(t)=14.0 A.
Steps for analysis:
- Initial condition at t=0t = 0t=0:
- The voltage across the inductor or resistor at t=0t = 0t=0 must be calculated from the function v0(t)v_0(t)v0(t). At t=0t = 0t=0, v0(0)=11.80−40e0=11.80−40=−28.20v_0(0) = 11.80 – 40 e^{0} = 11.80 – 40 = -28.20v0(0)=11.80−40e0=11.80−40=−28.20 V.
- The current i0(t)i_0(t)i0(t) is a constant after t=0t = 0t=0, so i0(t)=14.0i_0(t) = 14.0i0(t)=14.0 A for all t≥0t \geq 0t≥0.
- Exponential term interpretation:
- The term −40e−1000t-40 e^{-1000t}−40e−1000t suggests an exponentially decaying voltage over time, which is common in RL circuits as the inductor resists sudden changes in current.
- The rate of decay is determined by the time constant τ=LR\tau = \frac{L}{R}τ=RL, which controls how quickly the current or voltage approaches its steady-state value.
- Current through the inductor:
- For an RL circuit, the current is governed by the equation i(t)=i0(1−e−t/τ)i(t) = i_0 \left(1 – e^{-t/\tau} \right)i(t)=i0(1−e−t/τ), where i0i_0i0 is the final steady-state current.
- The current i0(t)i_0(t)i0(t) is given as constant, suggesting that the circuit is already in a steady state, and the voltage will approach a constant value after a certain time.
- Conclusion:
- The voltage decays over time, and the current remains constant after t=0t = 0t=0. This behavior is characteristic of RL circuits where the inductor initially opposes changes in current but eventually settles to a steady state.
Thus, v0(t)=11.80−40e−1000tv_0(t) = 11.80 – 40 e^{-1000t}v0(t)=11.80−40e−1000t and i0(t)=14.0i_0(t) = 14.0i0(t)=14.0 A represent the voltage and current for t≥0t \geq 0t≥0, respectively.
