Unit 2 (Chapter 3-4) Flashcards
What is the formula for Current in terms of Capacitance and Voltage ?
𝒾 = C d/dt[𝓋]
where
C is capacitance
d/dt[𝓋] is the derivative of the voltage function
What is formula for Voltage in terms of Capacitance and Current ?
𝓋(t) = 1/C ∫ 𝒾(t)dt + 𝓋₀
Formula for Capacitors in Parallel ?
Ceq = C1 + C2 + C3
Formula for Capacitors in Series ?
Ceq = 1/ [ (1/C1) + (1/C2) + (1/C3) ]
Formula for Inductances in Parallel ?
Formula for Inductances in Series ?
We are given a 5-μF capacitor that is charged
to 200 V. Determine the initial stored charge
and energy. If this capacitor is discharged to
0 V in a time interval of 1 μs, find the average
power delivered by the capacitor during the
discharge interval.
Initial Stored Energy: 𝓌(t) = ½C𝓋²(t)
𝓌(t) = ½(5-μF)(200V)²
𝓌(t) = 0.1J
Initial Stored Charge: 𝒾(t) = C𝓋(t)
𝒾(t) = (5-μF)(200V)
𝒾(t) = 1 mC
Find the voltage, power, and stored energy at
t = 10 ms for the capacitance in the circuit of
Figure P3.15.
Start with the function: 𝓋(t) = 1/C ∫ 𝒾(t)dt + 𝓋₀
𝓋(t) = 1/5μF ∫ (3mA)dt + 0 | t = [0,10]ms
𝓋(t) = (3mA / 5μF)t | t = [0,10]ms
𝓋(t) = 6V
Power is just: p(t) = 𝓋(t)𝒾(t)
p(t) = (6V)(3mA)
p(t) = 18mW
Energy Stored: 𝓌(t) = ½C𝓋²(t)
𝓌(t) = ½(5μF)(6V)²
𝓌(t) = 90μJ
Explain how to reduce this circuit
a) start with the top right Capacitors in series:
C = (2μF)(2μF) / 4μF
C = 1μF
Then use our new L to solve for the last part in Parallel
C = 1μF + 1μF
Ceq = 2μF
Find the equivalent capacitance between terminals x and y for circuit shown
in Figure P3.26.
Start by solving the Left Side
First two Capacitors are in Series:
CL = (1μF)(2μF) / 1μF + 2μF = 2/3μF
Now the part in Parallel:
CL = ⅔μF + 3μF
Solve the Right Side
First two Capacitors are in Parallel:
CR = 2μF
Now the second part is in Series:
CR = (2μF)(2μF) / 4μF = 1μF
Now the Capacitance Between the two terminals:
Cxy = CL + CR = 1μF + ⅔μF + 3μF
Cxy = 4⅔μF
Find the equivalent capacitance between terminals x and y for circuit shown
in Figure P3.26.
Since this is a waffle bridge we can split the circuit into two parallel parts.
Ct = 3μF
Cb = 6μF
Now we can solve the Capacitance between x and y in series:
Cxy = (6μF)(3μF) / (6+3)μF
Cxy = 2μF
Consider two initially uncharged capacitors
C1 = 15 μF and C2 = 10 μF connected in
series. Then, a 50-V source is connected to the
series combination, as shown in Figure P3.28.
Find the voltages v1 and v2 after the source is
applied. (Hint:The charges stored on the two
capacitors must be equal, because the current
is the same for both the capacitors.)
Start by solving the Circuit down and finding a value for the Current about the Circuit.
The Capacitors are in Series:
CEQ = (15μF)(10μF) / (25μF)
CEQ = 6μF
We can solve for the current about the Circuit now with our CEQ
𝒾 = C𝓋
𝒾 = (50V)(6μF) = 300μA
Now we can use this relation to solve for the Voltage in each Capacitor:
𝓋= 𝒾/C
𝓋1 = 300μA / 15μF
𝓋1 = 20V
𝓋2 = 300μA / 10μF
𝓋2 = 30V
A constant voltage of 10V is applied to a
50-μH inductance, as shown in Figure P3.45.
The current in the inductance at t = 0 is
−100-mA. At what time tₓ does the current
reach +100-mA?
All we need to do is use the equation: 𝓋(t) = Ld𝒾/dt
𝓋(t) = 10V
L = 50-μH
Δ𝒾 = 200-mA
Δt = tₓ
10V = (50-μH)(200-mA)/tₓ
tₓ = (50-μH)(200-mA) / 10V
tₓ = 1μs
What is the equation for Voltage across a Capacitor as a function of time ?
We treat the capacitor as an open circuit then solve for the Voltage across it
V𝒸(t) = V𝒸(∞) + [ V𝒸(0) - V𝒸(∞) ]e^(-t/Շ)
The circuit shown in Figure P4.26 is operating in steady state. Determine the values of 𝒾L, Vx, Vc
We start by setting t = ∞
This will stretch our Inductors into shorts and our Capacitors into opens
We can now do KVL to solve for Vx: Vx = 𝒾(R)
Vx = 15mA/3kohms = 45V
𝒾L will be be the original current value since we are open past the first resistor
Vc = -20V via KVL
Explain SOlution
We need to redraw in steady state.
Since the Resistor is isolated, we can solve directly as 𝒾R
𝒾R = 4mA
Now for the bottom we known through KVL the first considered Vc will be
Vc’(t) = 12V
Lastly we go back to the top where the resistor is found and solve for the voltage about the resistor:
Vc’‘(t) = 4mA(2kohms)
Vc’‘(t) = 8V
now the final Vc is the sum of the voltages:
Vc’‘(t) = 20V
Equation for Inductance 𝒾(t) at Steady State
𝒾(∞) = K1 + K2e^(-Rt/L)
