There is no energy stored in the circuit in the figure below at t = 0. Transform the circuit into the s domain and solve for Vo(s) and vo(t). 1 ΚΩ ww 0.2 μF 1 k vo(t) 0.2 µF 0.2 µF 100u(t)

In the following figure, the switch is left in position a for a long time interval and is then quickly thrown to position b. Find and rank the magnitudes of the voltages (V, VL, VRI, Vr2) across the four circuit elements a short time thereafter from the largest to the smallest. Vs 12.0 V 2.00 H 1 200 0 VRI 12.0 0 Vr2

ISMA 3K un 3K un + Vs 7mH 3√1₂ Im mn SmH -+ 20 V t IMF Vc What is Vs. iz Ve at steady-state?

Determine the inductor current i(t) for t20. Given that Vs = 10V, R₁ = 1kQ, R₂ = 5kQ, R3 = 2kQ2, and L = 1mH. 1=0 ww R₁LR₁

Determine the time constant of the circuit when the switch is throwninto position 1.Find the mathematical expression for the voltage across the capacitorand the current after the switch is thrown into position 1.Determine the magnitude of the voltage vC and the current iC the instantthe switch is thrown into position 2 at t = 1 s.

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The circuit in Figure has been connected for a long time. If the battery is disconnected, how does it take the capacitor to discharge to 1/20 of its initial voltage? Take R₁ = 2.00 Q, R₂=4.00 Q, R3-7.00 Q, R4 = 2.00 Q and C= 5.00 uF and ε = 14.0 V. (Your result must be in us. Include 2 digit after the decimal point and maximum of 2% of error is accepted in your answer.) E R₂ R₂ BA

Consider the circuit shown below, with an initially uncharged capacitor C and two identical resistors R. R O www V I C I₂ R C R At the instant the switch is closed, the current through the capacitor will be: zero At the instant the switch is closed, the current through the right-side resistor will be: V/R, directed Up After the switch has been closed for a very long time, it is opened. The current through the right-side resistor will be V/R, directed Down The switch is closed at t = 0. At t = ~ (long after) the current through the right-side resistor will be: V/2R, directed Down ✓ The switch is closed at t = 0. At t = ~ (long after) the current through the capacitor will be: V/2R, directed Up After the switch has been closed for a very long time, it is ope ed. The current through the be? V/R, directed Down pacitor will

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