Chapter 7 – Response of First Order RL and RC Circuits

Chapter 7 – Response of First Order RL and RC Circuits
Terminology Natural Response of an RL Circuit Natural Response of an RC Circuit The Step Response of RL and RC Circuits Sequential Switching Unbounded Response

Terminology First Order Circuit– Their voltages and currents are described by first order differential equations Natural Response – How the circuits respond when a switch is opened or closed

Natural Response of an RL Circuit
Inductor in terms of Thevenin and Norton Capacitor in terms of Thevenin and Norton

After a time, currents and voltages reach a constant value, when this happens, only dc currents can exist prior to the switch being opened. When the switch is closed, the inductor appears as a short, the voltage across the branch is zero. When L appears as a short, no current goes through R0 or R. The source current (𝐼 𝑆 ) all goes through L.

Deriving the expression for current: Take the differential with respect to time of both sides of the equation 𝑑𝑖 𝑑𝑡 𝑑𝑡=− 𝑅 𝐿 𝑖 𝑑𝑡 𝐿 𝑑𝑖 𝑑𝑡 +𝑅𝑖=0 𝑑𝑖 𝑖 =− 𝑅 𝐿 𝑑𝑡 𝑖( 𝑡 0 ) 𝑖(𝑡) 𝑑𝑥 𝑥 =− 𝑅 𝐿 𝑡 0 𝑡 𝑑𝑦 𝐿 𝑑𝑖 𝑑𝑡 =−𝑅𝑖 ln 𝑖 𝑡 − ln 𝑖 𝑡 0 =− 𝑅 𝐿 𝑡− 𝑡 𝑡 0 =0 𝑑𝑖 𝑑𝑡 =− 𝑅 𝐿 𝑖 𝑙𝑛 𝑖(𝑡) 𝑖(0) =− 𝑅 𝐿 𝑡 Apply base e to both sides 𝑒 𝑙𝑛 𝑖(𝑡) 𝑖(0) = 𝑒 − 𝑅 𝐿 𝑡 𝑖 𝑡 =𝑖(0) 𝑒 − 𝑅 𝐿 𝑡 𝑖(𝑡) 𝑖(0) = 𝑒 − 𝑅 𝐿 𝑡

𝑖 𝑡 =𝑖(0) 𝑒 − 𝑅 𝐿 𝑡 What is going on when 𝑡≥0

Instantaneous change of current cannot occur in an inductor The instant right after the switch is opened, the current in the inductor remains unchanged Use 0 − to denote time prior to switching Use to denote time immediately after switching 𝑖 0 − =𝑖 = 𝐼 0

𝑖 0 − =𝑖 = 𝐼 0 Substitute 𝑖 𝑡 =𝑖(0) 𝑒 − 𝑅 𝐿 𝑡 𝑖 𝑡 = 𝐼 0 𝑒 − 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥0 Current starts at and initial value 𝐼 0 and decreases exponentially toward zero as t increases

Voltage: 𝑣=𝑖𝑅= 𝐼 0 𝑅 𝑒 − 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥ 0 + 𝑣 0 − =0 𝑣 = 𝐼 0 𝑅 Voltage is only defined for 𝑡>0, not at 𝑡=0 because a step voltage occurs at 𝑡=0 Because of the step voltage the value at 𝑡=0 is unknown, therefore only 𝑡≥ 0 + defines the region of validity

Power: 𝑝= 𝐼 0 𝑒 − 𝑅 𝐿 𝑡 2 𝑅 𝑓𝑜𝑟 𝑡≥ 0 + Squaring the current yields: 𝑝= 𝐼 0 2 𝑅 𝑒 −2 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥ 0 +

Energy: 𝑤= 0 𝑡 𝑝𝑑𝑥 = 0 𝑡 𝐼 0 2 𝑅 𝑒 −2 𝑅 𝐿 𝑥 𝑑𝑥 =𝐼 0 2 𝑅 0 𝑡 𝑒 −2 𝑅 𝐿 𝑥 𝑑𝑥 =𝐼 0 2 𝑅 1 −2 𝑅 𝐿 𝑒 −2 𝑅 𝐿 𝑥 | 𝑡 0 = 𝐼 0 2 𝑅 1 −2 𝑅 𝐿 𝑒 −2 𝑅 𝐿 𝑡 − 𝑒 −2 𝑅 𝐿 0 = 𝐼 0 2 𝑅 1 −2 𝑅 𝐿 𝑒 −2 𝑅 𝐿 𝑡 −1 = 1 2 𝑅 𝐿 𝐼 0 2 𝑅 1− 𝑒 −2 𝑅 𝐿 𝑡 = 1 2 𝐿 𝐼 − 𝑒 −2 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥0

Recap: 𝑖 𝑡 = 𝐼 0 𝑒 − 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥0 𝑣= 𝐼 0 𝑅 𝑒 − 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥ 0 + 𝑝= 𝐼 0 2 𝑅 𝑒 −2 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥ 0 + 𝑤= 1 2 𝐿 𝐼 − 𝑒 −2 𝑅 𝐿 𝑡 𝑓𝑜𝑟 𝑡≥0

Time Constant – determines the rate at which the current or voltage approaches zero. 𝜏=𝑡𝑖𝑚𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡= 𝐿 𝑅 𝜏 −𝑇𝑎𝑢 Re-Write the equations with 𝜏 𝑖 𝑡 = 𝐼 0 𝑒 −𝑡 𝜏 𝑓𝑜𝑟 𝑡≥0 𝑣= 𝐼 0 𝑅 𝑒 −𝑡 𝜏 𝑓𝑜𝑟 𝑡≥ 0 + 𝑝= 𝐼 0 2 𝑅 𝑒 −2𝑡 𝜏 𝑓𝑜𝑟 𝑡≥ 0 + 𝑤= 1 2 𝐿 𝐼 − 𝑒 −2𝑡 𝜏 𝑓𝑜𝑟 𝑡≥0

Time Constant characteristics After 5 time constants, currents and voltages have reached their final values A long time- implies 5 or more time constants have elapsed Steady-state response – the response that exists a long time after switching has taken place The existence of current in the RL circuit is a momentary event and is called a transient response

At 5 time constants current is less than 1% of it’s initial value

Steps to calculate the natural response of an RL circuit Find the initial current, 𝐼 0 , through the inductor. Find the time constant of the circuit, 𝜏= 𝐿 𝑅 . Use 𝑖 𝑡 = 𝐼 0 𝑒 −𝑡 𝜏 to calculate 𝑖(𝑡).

Example: The switch has been closed for a long time before it is opened at t = 0. Find 𝑖 𝐿 𝑡 , 𝐼 0 𝑡 , 𝑣 0 𝑡 , and the percentage of the total energy stored in the 2H inductor that is dissipated in the 10Ω resistor. Since the switch has been closed a long period of time, the inductor appears as a short, concluding that the initial current through the inductor is 20A. 𝑅 𝑒𝑞 = − −1 −1 =10Ω 𝜏= 𝐿 𝑅 𝑒𝑞 = 2 10 =0.2 𝑖 𝐿 𝑡 = 𝐼 0 𝑒 − 𝑡 𝜏 =20 𝑒 − 𝑡 0.2 =20 𝑒 −5𝑡 𝐴 The equivalent resistance across the terminals of the inductor will be 𝑖 𝐿 𝑡 =20 𝑒 −5𝑡 𝐴

𝑖 𝐿 𝑡 =20 𝑒 −5𝑡 𝐴 Now calculate 𝑖 0 Using a current divider: 𝑖 0 (𝑡)=− 𝑖 𝐿 =−20 𝑒 −5𝑡 =−4 𝑒 −5𝑡 𝑖 0 (𝑡)=−4 𝑒 −5𝑡 𝐴

Using Ohms law: 𝑣 0 = 𝑖 0 𝑅 Previously solved 𝑖 0 =−4 𝑒 −5𝑡 𝐴 𝑣 0 (𝑡)=(−4 𝑒 −5𝑡 )(40) 𝑣 0 𝑡 =−160 𝑒 −5𝑡

Find the percentage of the total energy stored in the 2H inductor that is dissipated in the 10Ω resistor. Previously 𝑣 0 𝑡 =−160 𝑒 −5𝑡 𝑉 𝑝 10Ω 𝑡 = 𝑣 = −160 𝑒 −5𝑡 =2560 𝑒 −10𝑡 𝑊 Total energy dissipated in the 10Ω resistor: 𝑤 10Ω 𝑡 = 0 ∞ 𝑝 𝑑𝑡 = 0 ∞ 2560 𝑒 −10𝑡 𝑑𝑡 = 2560 −10 𝑒 −10𝑡 | ∞ 0 = 2560 −10 𝑒 −10(∞) − 𝑒 −10(0) = 2560 −10 0−1 =256𝐽

Previously 𝑤 10Ω 𝑡 =256J Initial energy stored in the 2H inductor 𝑤 0 = 1 2 𝐿 𝑖 2 0 = =400𝐽 The percentage of energy dissipated in the 10Ω resistor is: =64%

Natural Response of an RC Circuit
Assume the switch has been in a position for a long time to reach steady state Capacitor behaves as an open in the presence of a constant voltage Voltage source cannot sustain a current so a voltage appears across the capacitor terminals (appears open)

Expressions for tau, voltage, current, power and energy for a capacitor: 𝜏=𝑅𝐶 𝑣 𝑡 = 𝑉 0 𝑒 − 𝑡 𝜏 𝑓𝑜𝑟 𝑡 ≥0 𝑖 𝑡 = 𝑣(𝑡) 𝑅 = 𝑉 0 𝑅 𝑒 − 𝑡 𝜏 𝑓𝑜𝑟 𝑡≥ 0 + 𝑝=𝑣𝑖= 𝑉 𝑅 𝑒 − 2𝑡 𝜏 𝑓𝑜𝑟 𝑡≥ 0 + 𝑤= 0 𝑡 𝑝 𝑑𝑥 = 0 𝑡 𝑉 𝑅 𝑒 − 2𝑡 𝜏 𝑑𝑥= 1 2 𝐶 𝑉 − 𝑒 − 2𝑡 𝜏 𝑓𝑜𝑟 𝑡≥0

Steps for calculating the natural response of an RC circuit: Find the initial voltage 𝑉 0 across the capacitor. Find the time constant 𝜏=𝑅𝐶. Calculate 𝑣 𝑡 = 𝑉 0 𝑒 − 𝑡 𝜏 using 𝑉 0 and 𝜏.

Example: The switch in the circuit has been in position x for a long time. At t=0, the switch moves instantaneously to position y. Calculate 𝑣 𝐶 𝑡 , 𝑣 0 𝑡 , 𝑖 0 𝑡 and the total energy dissipated in the 60kΩ resistor. First, calculate 𝑣 𝐶 (𝑡) Because the switch has been closed a long time, the capacitor will charge to 100V So 𝑉 0 =100𝑉 𝑅 𝑒𝑞 = 240𝑘 −1 + 60𝑘 −1 −1 +32𝑘=80𝑘Ω 𝜏=𝑅𝐶= 80 × × 10 −6 =40𝑚𝑠 Substitute in: 𝑣 𝐶 𝑡 = 𝑉 0 𝑒 − 𝑡 𝜏 =100 𝑒 − 𝑡 =100 𝑒 −25𝑡

Find 𝑣 0 (𝑡) using a voltage divider Previously 𝑅 𝑒𝑞 = 240𝑘 −1 + 60𝑘 −1 −1 =80𝑘Ω 𝑣 𝐶 𝑡 =100 𝑒 −25𝑡 𝑣 0 𝑡 = 𝑣 𝐶 𝑡 = 𝑒 −25𝑡 =60 𝑒 −25𝑡 𝑉 Total parallel resistance equals 48kΩ 𝑅 𝑒𝑞 = 240𝑘 −1 + 60𝑘 −1 −1

Calculate 𝑖 0 (𝑡) using Ohm’s Law Previously 𝑣 0 𝑡 =60 𝑒 −25𝑡 𝑉 𝑖 0 𝑡 = 𝑣 0 (𝑡) 𝑅 = 60 𝑒 −25𝑡 60× = 𝑒 −25𝑡 𝑚𝐴

Calculate the total energy dissipated in the 60kΩ resistor. Previously 𝑣 0 𝑡 =60 𝑒 −25𝑡 𝑉 𝑖 0 𝑡 = 𝑒 −25𝑡 𝑚𝐴 𝑝= 𝑖 0 (𝑡) 2 𝑅= 𝑒 −25𝑡 = 60𝑒 −50𝑡 𝑚𝑊 𝑤 60𝑘Ω = 0 ∞ 𝑝 𝑑𝑡 = 0 ∞ 𝑒 −50𝑡 𝑑𝑡 = −50 𝑒 −50𝑡 | ∞ 0 = −50 𝑒 −50 ∞ − 𝑒 − = −50 (0−1) =1.2𝑚𝐽

Step Response of an RL Circuit
When dealing with RL step response, find the expression for current in the circuit and the voltage across the inductor 𝑖 𝑡 = 𝑉 𝑠 𝑅 + 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 − 𝑅 𝐿 𝑡 Step Response of RL circuit 𝑖 𝑡 = 𝑉 𝑠 𝑅 + − 𝑉 𝑠 𝑅 𝑒 − 𝑅 𝐿 𝑡 When initial energy is zero, 𝐼 0 =0 therefore

Step response voltage of an RL circuit Previously 𝑣=𝐿 𝑑𝑖 𝑑𝑡 𝑖 𝑡 = 𝑉 𝑠 𝑅 + 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 − 𝑅 𝐿 𝑡 𝑣=𝐿 − 𝑅 𝐿 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 − 𝑅 𝐿 𝑡 𝑣= −𝑅 𝐼 0 + 𝑉 𝑠 𝑒 − 𝑅 𝐿 𝑡 When initial inductor current is zero 𝑣= 𝑉 𝑠 𝑒 − 𝑅 𝐿 𝑡

Example: The switch has been in a position for a long time. At t=0, the switch moves from position a to position b. The switch is a make-before-break type, which means the connection at position b is established before the connection at position a is broken so there is no interruption of the current through the inductor Find i(t), v(t), find when the voltage is 24V, plot i(t) and v(t) 𝑖 𝑡 = 𝑉 𝑠 𝑅 + 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 − 𝑅 𝐿 𝑡 Current is orientated opposite to the reference direction I, so 𝐼 0 =−8𝐴 𝑖 𝑡 = −8− 𝑒 − 𝑡 𝑖 𝑡 =12−20 𝑒 −10𝑡 𝐴

𝑣=𝐿 𝑑𝑖 𝑑𝑡 Previously 𝑖 𝑡 =12−20 𝑒 −10𝑡 𝐴 𝑑𝑖 𝑑𝑡 =200 𝑒 −10𝑡 𝑣= 𝑒 −10𝑡 𝑣=40 𝑒 −10𝑡 The initial inductor voltage is 𝑣 =40 𝑒 −10(0) =40𝑉

Finding the time when the inductor voltage is 24V Previously 𝑣=40 𝑒 −10𝑡 24=40 𝑒 −10𝑡 24 40 = 𝑒 −10𝑡 − =−10𝑡 Divide both sides by 40 =𝑡 0.6= 𝑒 −10𝑡 𝑡=51.08 𝑚𝑠 ln 0.6 = ln 𝑒 −10𝑡 Take the natural log of both sides

Graphing i(t) and v(t) 𝑖 𝑡 =12−20 𝑒 −10𝑡 𝐴 𝑣=40 𝑒 −10𝑡 𝑡=51.08 𝑚𝑠 𝑣=40𝑉

Step Response of an RC Circuit
𝑣 𝐶 = 𝐼 𝑠 𝑅+ 𝑉 0 − 𝐼 𝑠 𝑅 𝑒 − 𝑡 𝑅𝐶 𝑖= 𝑖 𝑠 − 𝑉 0 𝑅 𝑒 − 𝑡 𝑅𝐶

Example: The switch has been in position 1 for a ling time. At t=0, the switch moves to position 2. Find 𝑣 0 (𝑡) and 𝑖 0 (𝑡). Since the switch has been in position 1 for a long time, use a voltage divider to find the initial value of 𝑣 0 Now short the -75V source to calculate the Thevenin Resistance 𝑣 0 = =30𝑉 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑅 𝑡ℎ = − −1 −1 =40𝑘Ω 𝑁𝑜𝑟𝑡𝑜𝑛 𝑐𝑢𝑟𝑟𝑒𝑛𝑡= 𝑉 𝑜𝑝𝑒𝑛𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑅 𝑡ℎ = − =−1.5𝑚𝐴 To find the voltage for 𝑡≥0, find the open circuit voltage across the capacitor to find a Thevenin and Norton equivalence. Using a voltage divider 𝑉 𝑜𝑝𝑒𝑛𝑐𝑖𝑟𝑐𝑢𝑖𝑡 = −75 =−60 𝑉

Previously 𝑣 0 = 𝐼 𝑠 𝑅+ 𝑉 0 − 𝐼 𝑠 𝑅 𝑒 − 𝑡 𝑅𝐶 𝑉 𝑜𝑝𝑒𝑛 𝑐𝑖𝑟𝑐𝑢𝑡 = 𝐼 𝑠 𝑅=−60𝑉 𝑉 0 =30𝑉 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑅𝐶= 𝑅 𝑡ℎ 𝐶= × 10 −6 =10𝑚𝑠 Substitute values into 𝑣 0 𝑣 0 =−60+ 30− −60 𝑒 − 𝑡 .010 𝑣 0 =−60+90 𝑒 −100𝑡 𝑉 𝑓𝑜𝑟 𝑡≥0

Previously 𝑖= 𝑖 𝑠 − 𝑉 0 𝑅 𝑒 − 𝑡 𝑅𝐶 𝑉 0 =30𝑉 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑅𝐶=10𝑚𝑠 𝑁𝑜𝑟𝑡𝑜𝑛 𝑐𝑢𝑟𝑟𝑒𝑛𝑡= 𝑖 𝑠 = 𝑉 𝑜𝑝𝑒𝑛𝑐𝑖𝑟𝑐𝑢𝑖𝑡 𝑅 𝑡ℎ = − =−1.5𝑚𝐴 Substitute in values 𝑖 0 = 𝑖 𝑠 − 𝑉 0 𝑅 𝑒 − 𝑡 𝑅𝐶 𝑖 0 = −.0015− 𝑒 − 𝑡 .010 𝑖 0 =−2.25 𝑒 −100𝑡 𝑚𝐴

General solution for natural and step responses of RL and RC Circuits
𝑥 𝑡 = 𝑥 𝑓 + 𝑥 𝑡 0 − 𝑥 𝑓 𝑒 (−𝑡+ 𝑡 0 ) 𝜏 [ ] Unknown variable as a function of time The final value of the variable The initial value of the variable The final value of the variable = + - × 𝑒 −[𝑡− 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑠𝑤𝑖𝑡𝑐ℎ𝑖𝑛𝑔 ] (𝑡𝑖𝑚𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) Identify the variable of interest. Capacitive voltage for RC circuits and inductive current for RL circuits. Determine the initial value of the variable. Calculate the final value of the variable. Calculate the time constant for the circuit.

General solution for a step response of an RC Circuit
Example: The switch has been in a position for a long time. At t=0 the switch is moved to position b. Find the initial value of 𝑣 𝐶 , the final value of 𝑣 𝐶 , the time constant, the expression for 𝑣 𝐶 (𝑡), the expression for 𝑖(𝑡), and how long after the switch is in position b does the capacitor voltage equal zero. Initial value of 𝑣 𝐶 is the same voltage that is across the 60Ω resistor because the switch was in position a. Use a voltage divider 𝑣 𝐶 = =30𝑉 Since the reference for 𝑣 𝐶 is positive at the upper terminal of the capacitor, 𝑣 𝐶 initial = -30 V Final value of 𝑣 𝐶 , after the switch has been in position b for a long time, the capacitor appears as an open in terms of the 90V source. The capacitor voltage can never exceed the source voltage.

Time constant 𝜏=𝑅𝐶= × 10 −6 =0.2𝑠 Previously 𝑣 𝐶 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =−30𝑉 𝑣 𝐶 𝑓𝑖𝑛𝑎𝑙 =90𝑉 𝑥 𝑡 = 𝑥 𝑓 + 𝑥 𝑡 0 − 𝑥 𝑓 𝑒 (−𝑡+ 𝑡 0 ) 𝜏 𝑣 𝐶 𝑡 = 𝑣 𝐶𝑓𝑖𝑛𝑎𝑙 + 𝑣 𝐶 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑣 𝐶𝑓𝑖𝑛𝑎𝑙 𝑒 (−𝑡+ 𝑡 0 ) 𝜏 𝑣 𝐶 𝑡 =90+(−30−90) 𝑒 −5𝑡 𝑣 𝐶 𝑡 =90−120 𝑒 −5𝑡

Previously 𝜏=0.2𝑠 𝑣 𝐶 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =−30𝑉 Find the initial and final current values The current in the capacitor can change instantaneously. The current is equal to the current in the 400kΩ resister. 𝑖 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = [90− −30 ] =300𝜇𝐴 In a capacitor, the voltage cannot change instantaneously, therefore 𝑖 𝑓𝑖𝑛𝑎𝑙 =0 𝑖 𝑡 = 𝑖 𝑓𝑖𝑛𝑎𝑙 + 𝑖 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑖 𝑓𝑖𝑛𝑎𝑙 𝑒 (−𝑡+ 𝑡 0 ) 𝜏 𝑖 𝑡 =0+[300𝜇𝐴−0] 𝑒 −5𝑡 𝑖 𝑡 =300 𝑒 −5𝑡 𝜇𝐴

Previously 𝑣 𝐶 𝑡 =90−120 𝑒 −5𝑡 How long after the switch is in position b does the capacitor voltage equal zero? Set 𝑣 𝐶 𝑡 =0 then solve for t 0.75= 𝑒 −5𝑡 0=90−120 𝑒 −5𝑡 ln 0.75 = ln 𝑒 −5𝑡 90=120 𝑒 −5𝑡 − =−5𝑡 = 𝑒 −5𝑡 57.54𝑚𝑠=𝑡

𝑣 𝐶 𝑡 =90−120 𝑒 −5𝑡 𝑖 𝑡 =300 𝑒 −5𝑡 𝜇𝐴 𝑣 𝐶 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =−30𝑉 𝑖 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 =300𝜇𝐴

General solution for a step response of an RL Circuit
Example: The switch has been open for a long time. At t=0, the switch is closed. Find the equation for v(t) and i(t) The switch has been open for a long time, the initial current is 𝐼 0 = =5𝐴 When the switch is closed, it creates a short that bypasses the 3Ω resistor. The initial voltage across the inductor becomes 𝑣 0 =20−5 1 =15𝑉 The time constant is 𝜏= 𝐿 𝑅 = 80 1 =80𝑚𝑠 𝑣 𝑓𝑖𝑛𝑎𝑙 =0𝑉 Substitute into the general solution 𝑣 𝑡 =0+(15−0) 𝑒 −𝑡 80𝑚𝑠 𝑣 𝑡 =15 𝑒 −12.5𝑡 𝑉

General solution for a step response of an RL Circuit
Previously 𝐼 0 =5𝐴 𝜏=80𝑚𝑠 After the switch has been closed a long time, the final current through the inductor becomes 𝐼 𝑓𝑖𝑛𝑎𝑙 = 20 1 =20𝐴 Substitute into the general solution 𝑖 𝑡 =20+(5−20) 𝑒 −𝑡 80𝑚𝑠 𝑖 𝑡 =20−15 𝑒 −12.5𝑡 A

Sequential Switching Sequential switching – when switching occurs more than once in a circuit Example: The two switches have been closed for a long time. At t=0, switch 1 is opened. Then, 35ms, later, switch 2 is opened. Find 𝑖 𝐿 (𝑡) for 0≤𝑡≤35𝑚𝑠 Find 𝐼 0 first for when t < 0. Inductor acts as a short. 𝑅 𝑇 = − −1 + 3 −1 −1 −1 =5.71Ω Use a current divider 𝑖 4Ω = =10.5𝐴 𝑖 2Ω = =9𝐴 𝑖 3Ω = 𝑖 𝐿 0 − Use a current divider 𝑖 𝐿 0 − = 𝐴=6𝐴 𝜏= 𝐿 𝑅 𝐿 = =0.025 𝑜𝑟 25𝑚𝑠 𝑅 𝐿 = − −1 −1 =6Ω

Sequential Switching Previously 𝑖 𝐿 0 − =6𝐴 𝜏=25𝑚𝑠
𝑖 𝐿 0 − =6𝐴 𝜏=25𝑚𝑠 𝑖 𝐿 𝑡 =6 𝑒 −𝑡 =6 𝑒 −40𝑡 𝐴 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠 Now find 𝑖 𝐿 𝑡 for 35ms≤𝑡 Substitute 35ms in for t 𝑖 𝐿 𝑡 =6 𝑒 −40(.035) =1.479𝐴 When switch 2 is opened at 35ms the circuit reduces to 𝜏= 𝐿 𝑅 𝐿 = =0.016=16.6𝑚𝑠 𝑅 𝐿 =9Ω 𝑖 𝐿 =1.479 𝑒 − 𝑡− 𝜏 =1.479 𝑒 − 𝑡− =1.479 𝑒 −60(𝑡−0.035) Represents after 35ms

Sequential Switching What percentage of energy is stored in the 150mH inductor is dissipated in the 18Ω resistor? Previously 𝑖 𝐿 𝑡 =6 𝑒 −40𝑡 𝐴 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠 𝑣 𝐿 =𝐿 𝑑𝑖 𝑑𝑡 =0.15 𝑑 𝑑𝑡 6 𝑒 −40𝑡 Initial energy in the inductor: 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠 𝑣 𝐿 = 𝑒 −40𝑡 −40 =−36 𝑒 −40𝑡 𝑉 𝑤 0 = 1 2 𝐿 𝑖(0) 2 𝑝 18Ω = 𝑣 𝐿 𝑅 18Ω = −36 𝑒 −40𝑡 =72 𝑒 −80𝑡 𝑤 0 = 1 2 (0.15) 6 𝑒 −40𝑡 2 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠 𝑤 0 = 1 2 (0.15) 6 𝑒 −40(0) 2 𝑤 18Ω = 𝑒 −80𝑡 𝑑𝑡 = 72 −80 𝑒 −80𝑡 | 𝑤 0 = =2.7𝐽 =0.9 1− 𝑒 −2.8 =845.27𝑚𝐽

Sequential Switching What percentage of energy is stored in the 150mH inductor is dissipated in the 18Ω resistor? Previously 𝑤 0 =2.7𝐽 𝑤 18Ω =845.27𝑚𝐽 𝑣 𝐿 = −36 𝑒 −40𝑡 845.27𝑚𝐽 2700𝑚𝐽 =0.3131=31.31% 2.7J = 2700mJ What percentage of energy is stored in the 150mH inductor is dissipated in the 3Ω resistor? 𝑝 3Ω = 𝑣 3Ω 𝑅 3Ω = −12 𝑒 −40𝑡 =48 𝑒 −80𝑡 𝑣 3Ω =𝑖𝑅 𝑤 3Ω = 𝑒 −80𝑡 𝑑𝑡 = 48 −80 𝑒 −80𝑡 | 𝑣 3Ω = 𝑣 𝐿 (3) =0.6 1− 𝑒 −2.8 =563.51𝑚𝐽 𝑣 3Ω = 𝑣 𝐿 3 = −36 𝑒 −40𝑡 3 =−12 𝑒 −40𝑡 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠

Sequential Switching What percentage of energy is stored in the 150mH inductor is dissipated in the 3Ω resistor? Previously 𝑤 0 =2.7𝐽 𝑣 𝐿 = −36 𝑒 −40𝑡 𝑓𝑜𝑟 0≤𝑡≤35𝑚𝑠 𝑤 3Ω =563.51𝑚𝐽 Now find the energy for the 3Ω for when 𝑡≥35𝑚𝑠 Use 𝑖 𝐿 previously found to be 𝑖 𝐿 =1.479 𝑒 −60(𝑡−0.035) that is the current through the 3Ω resistor 𝑝 3Ω = 𝑖 𝐿 2 𝑅 3Ω = 𝑒 −60 𝑡− =2.187 𝑒 −120 𝑡− (3) 𝑤 3Ω = ∞ 𝑝 𝑑𝑡 = ∞ 𝑒 −120(𝑡−0.035) 𝑑𝑡 = −120 𝑒 −120(𝑡−0.035) | ∞ = −120 0−1 = =54.68𝑚𝐽 𝑤 3Ω 𝑡𝑜𝑡𝑎𝑙 = = 𝑚𝐽 Percentage of energy stored is ×100=22.9%

Sequential Switching What percentage of energy is stored in the 150mH inductor is dissipated in the 6Ω resistor? Since the 6Ω resistor is in series with the 3Ω resistor, the energy dissipated will be twice that of the 3Ω 𝑤 6Ω = = 𝑚𝐽 Percentage of energy stored is ×100=45.79%

Unbounded Response Any time that a circuit response grows rather than decays is called an unbounded response Because of dependent sources, the Thevenin resistance may be negative with respect to the terminals of an inductor or capacitor Use the test source method to find the Thevenin 𝑖 ∆ = 𝑣 𝑇 20 𝑖 𝑇 = 𝑣 𝑇 − 1 5 𝑚𝐴 𝑖 𝑇 = 𝑣 𝑇 10 −7 𝑣 𝑇 𝑣 𝑇 20 𝑚𝐴 𝑣 𝑇 𝑖 𝑇 =−5𝑘Ω= 𝑅 𝑇ℎ 𝑖 𝑇 = 𝑣 𝑇 − 𝑚𝐴

Unbounded Response 5×10 −6 𝑑 𝑣 0 𝑑𝑡 − 𝑣 0 5𝑘Ω =0 𝑑 𝑣 0 𝑑𝑡 −40 𝑣 0 =0
5×10 −6 𝑑 𝑣 0 𝑑𝑡 − 𝑣 0 5𝑘Ω =0 𝑑 𝑣 0 𝑑𝑡 −40 𝑣 0 =0 Separation of variables 𝑑 𝑣 0 𝑑𝑡 =40 𝑣 0 Assume the capacitor short-circuits when its terminal voltage reaches 150V. How many milliseconds elapse before the capacitor short circuits? 𝑒 ln 𝑣 0 (𝑡) 𝑣 0 = 𝑒 40𝑡 𝑑 𝑣 0 𝑣 0 =40𝑑𝑡 150=10 𝑒 40𝑡 V 𝑣 0 (𝑡) 𝑣 0 = 𝑒 40𝑡 𝑑 𝑣 0 𝑣 0 = 40𝑑𝑡 15= 𝑒 40𝑡 V 𝑣 0 𝑡 = 𝑣 0 𝑒 40𝑡 ln 15 = ln 𝑒 40𝑡 V ln 𝑣 0 (𝑡) 𝑣 0 +𝐶=40𝑡+𝐶 2.7080=40𝑡 𝑣 0 𝑡 =10 𝑒 40𝑡 V =𝑡 𝑡=67.7 𝑚𝑠

Chapter 7 – Response of First Order RL and RC Circuits

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Chapter 7 – Response of First Order RL and RC Circuits

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Presentation on theme: "Chapter 7 – Response of First Order RL and RC Circuits"— Presentation transcript:

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