CPEN 206 Linear Circuits Lecture # 3 Circuit Configurations Dr. Godfrey A. Mills Email: gmills@ug.edu.gh Phone: 0269073163 February 15, 2016 Course TA David S. Tamakloe CPEN 206 Lecture 3 2015_2016 1
Circuit Configurations o Passive circuit elements in electrical network are usually arranged in three configurations. These are n Series circuit configuration n Parallel circuit configuration n Hybrid circuit configuration o Generally, practical circuits we encounter are quite complicated and for easy analysis, the circuits are replaced by their equivalent forms. o Reducing complicated circuits to the simplified form is known as equivalent circuit. o Complicated networks of resistors or impedance in either series or parallel form can be replaced by a single equivalent resistor or impedance. CPEN 206 Lecture 3 2015_2016 2
Circuit reduction approach o To reduce a circuit to its simplest form, we use the reduction techniques below: n Assess which elements (resistors or impedances) in the circuit are connected in series or parallel n Redraw the circuit, replacing all equivalent of the series or parallel element identified in step # 1 n Repeat steps # 1 and 2 until the entire circuit is reduced to a single equivalent element n If the active source voltage is known, then we can calculate the total current from the total voltage and total value of the equivalent circuit element CPEN 206 Lecture 3 2015_2016 3
Circuit reduction approach o Cont d.: n Taking the total current and voltage values, go back to the last step in the circuit reduction process and insert those values where applicable n From the known total equivalent value of the circuit element and total voltage or current values from the above step, use Ohms law to calculate the unknown values of voltage and current n Repeat the last two steps until all voltage and current values in the original circuit configuration are known n Calculate the power dissipation from known voltage, current and circuit elements (resistance or impedance) values CPEN 206 Lecture 3 2015_2016 4
Equivalent Resistance i(t) v(t) i(t) R eq v(t) o R eq is equivalent to the resistor network on the left in the sense that they have the same iv characteristics. o Rest of the circuit cannot tell whether the resistor network or the equivalent resistor is connected to it. o Equivalent resistance cannot be used to find voltages or currents internal to the resistor network. CPEN 206 Lecture 3 2015_2016 5
Series Circuits o Series resistances involves arrangement of joining resistances in endtoend form to have a single path for electrons to flow. o This can be represented by equivalent resistance to make the circuit simpler and easier to solve, and also make it easier to understand. o Being a series circuit: n Current is the same through all the resistors n Voltage across each resistor is different due to its resistance and is given by OHM S LAW. n Sum of all voltage drops is equal to the applied voltage all three resistances CPEN 206 Lecture 3 2015_2016 6
Series Impedance Z 1 Z 3 Z 2 Z eq Z eq = Z 1 Z 2 Z 3 CPEN 206 Lecture 3 2015_2016 7
Series Characteristics o Same current flows through all parts of the circuit o Different resistors have individual voltage drops o Voltage drops in the circuit are additive o Applied source voltage is always equal to the sum of different voltage drops in the circuit o Resistances (or elements) in the circuit are additive o Powers dissipated in the elements in the circuit are additive CPEN 206 Lecture 3 2015_2016 8
Series Example 1 q Three resistors R1, R2, and R3 are connected in series across a 12V battery as illustrated in the circuit diagram below. The first resistor R1 has a value of 1Ω, the second resistor R2 has a voltage drop of 4V across its terminals and the third resistor R3 has a power dissipation of 12W. Calculate the value of the circuit current I that is delivered by the source. 1 2 3 4 I V 1 V 2 V 3 V CPEN 206 Lecture 3 2015_2016 9
Series Solution 1 o o o o o o First step is to identify what we have n R1 = 1.0, IR2 = 4, and I 2 R3 = 12 From above, we derive expression between R2 and R3 as n I 2 = 16/R 2 2 and I2 = I 2 /R3 >> R3 = (3/4)R 2 2 Since the resistors are serially arranged, the current is: I(R1 R2 R3) = 12 Substituting the values for I and R3 in above gives n 3R 2 2 8R2 4 = 0 Solving the above equation gives two solutions for R2 as n R2 = 2Ω or R2 = (2/3) Ω and R3 = 3Ω or R3 = (1/3) Ω Substituting values of R1, R2, and R3, gives current as n I = 2A or I = 6A CPEN 206 Lecture 3 2015_2016 10
Practice Questions 1 o Question 1: Three resistor 100Ω, 200Ω, and 300Ω are connected in series to a 220V battery. Draw the circuit diagram and find the current in the circuit and power dissipated in each resistor. o Question 2: A 20Ω resistor is connected in series with an unknown resistor R, and the two are connected across a 220V battery. Draw the circuit diagram. If the power loss in R is 50W, find the value of R. o Question 3: A certain device is rated at 120V, 100W. Find the resistor that will be required to reduce the voltage of the line from 120V to 80V for the operation of the device. CPEN 206 Lecture 3 2015_2016 11
Practice Questions 2 o Question 1: A passive lowpass filter circuit comprising an R and C, are arranged in a serial configuration. The voltage source to the circuit is Vs. If the output voltage measured across the capacitor is Vo, use the voltage division concept to derive an expression for the output voltage Vo. Find ratio of the output voltage to the input source voltage, thus, (Vo/Vs). o Question 2: If the position of the filter elements are changed in the above lowpass filter (C and R), we have a highpass. If the output voltage is measured across the resistor instead of the capacitor, use the voltage division method to derive an expression for the ratio of the output voltage to the input voltage, thus, (Vo/Vs). CPEN 206 Lecture 3 2015_2016 12
Practice Questions 3 o Question 3: Assume the passive filter above is now redesigned to serve as a bandpass filter circuit comprising an L, C, and R, all arranged in series. If the input voltage to the circuit is Vs and the output voltage measured across the resistor only is Vo, use the voltage division method to derive an expression for the output voltage Vo. Find ratio of the output voltage to the input voltage, thus, (Vo/Vs). o Question 4: Now, assume the filter circuit above is now designed to serve as a bandstop filter, where the output voltage Vo is measured across the combined C and L. Derive an expression for the output voltage and find ratio of the output voltage to the input voltage, thus, (Vo/Vs). CPEN 206 Lecture 3 2015_2016 13
Parallel Circuits o Circuits are said to be in parallel when two or more elements are connected to the same voltage source in such a way that there is the creation of multiple paths with common electrically points for electrons to flow from one end of the battery to the other. o If these elements are resistances; R1, R2, R3,,,,Rn, for n resistance, connected in parallel with the supply voltage, then the current through the various resistances will be n I 1 = V/R 1, I 2 = V/R 2,,,,,, I n = V/R n o Total current through the source will be sum of currents n I = I 1 I 2,,,,,,I n n V/R 1 V/R 2,,,,, V/R n CPEN 206 Lecture 3 2015_2016 14
Parallel Circuits o This gives us n I/V = 1/R1 1/R2,,,,, 1/Rn o This can be represented by an equivalent resistance to make the circuit simpler, easier to solve, and easier to understand. o We see that for parallel circuit: n The potential difference across all three resistances is the same n The current through each resistance is different due to its different resistance and is given by OHM S LAW. n The total current is equal to the sum of all three separate currents CPEN 206 Lecture 3 2015_2016 15
Parallel Impedance Z 1 Z 2 Z 3 Z eq 1/Z eq = 1/Z 1 1/Z 2 1/Z 3 CPEN 206 Lecture 3 2015_2016 16
Parallel Example 1 n Question : n Two coils are connected in parallel and a voltage of 200V is applied to the terminals. The total current delivered by the source is 25A and the power dissipated in one of the coils is 1500W. Determine the resistance of each coil and the currents through each coil. n Solution : n Since P in one coil = 1500W and V = 200V, we have R = V 2 /R = 26.67Ω. n Current I = 200/26.67 = 7.5A. n Since total current is 25A, current through the other coil is 25A 7.5A = 17.5A. n Resistance of other coil = 200/17.5 = 11.42Ω. CPEN 206 Lecture 3 2015_2016 17
Practice Questions 4 o Question 1: Three resistances 750Ω, 600Ω, and 200Ω are connected in parallel. If the total current is 1A, find the voltage applied to the terminals and the current in each branch. o Question 2: Three coils R1, R2, and R3 are connected in parallel and a voltage of 12V is applied to the terminals. If R1 = 3Ω, R2 = 6Ω, and R3 = 9Ω, find (a) the total current delivered by the source, and (b) the current through each coil. CPEN 206 Lecture 3 2015_2016 18
Hybrid Circuits o The combination of series and parallel elements of resistance or impedances can be used to find voltages and currents in circuits. o This process can often yield the fastest solutions to networks. The process may not apply to complicated networks. o Resistance or impedances are combined to create a simple circuit (usually one source and one resistance), from which a voltage or current can be found o Once this is done, the voltage or current in the network can easily be found. CPEN 206 Lecture 3 2015_2016 19
Hybrid Circuits i(t) v(t) i(t) R eq v(t) o R eq is equivalent to the resistor network on the left in the sense that they have the same iv characteristics. o The rest of the circuit cannot tell whether the resistor network or the equivalent resistor is connected to it. o The equivalent resistance cannot be used to find voltages or currents internal to the resistor network. CPEN 206 Lecture 3 2015_2016 20
Hybrid Example 1 CPEN 206 Lecture 3 2015_2016 21
Hybrid Example 2 o Ladder networks are used in analogtodigital converters to provide reference voltages that are 1/2, 1/4, 1/8, etc. of a source voltage. 2kΩ 2kΩ o Find the equivalent resistance of the ladder network shown in the circuit above. CPEN 206 Lecture 3 2015_2016 22
Hybrid Solution 2 o Find the equivalent resistance by making combinations of series and parallel resistors until you have only one resistor left. 1k 1k 2k 2k 2k 2k 1k CPEN 206 Lecture 3 2015_2016 23
Hybrid Solution 2 2kΩ 2kΩ o The equivalent resistance of the ladder network is. CPEN 206 Lecture 3 2015_2016 24
Hybrid Example 3 10Ω 769pF 159µH o The circuit diagram above is a bandpass filter network. If the angular frequency, ω = 2.86 10 6, find the equivalent impedance of the circuit. CPEN 206 Lecture 3 2015_2016 25
Hybrid Solution 3 j455ω 10Ω j455ω o Now combine series impedances CPEN 206 Lecture 3 2015_2016 26
Hybrid Solution 3 j455ω 10Ω j455ω 455.1Ω 88.7 o Now combine parallel impedances CPEN 206 Lecture 3 2015_2016 27
Hybrid Solution 3 ( j455) j455 10 j455 10 j455 = 20.7kΩ 1.3 o The final results expressed in phasor form. The answer can also be reduced to a complex rectangular form CPEN 206 Lecture 3 2015_2016 28
Hybrid Example 4 10Ω 769pF 159µH 50kΩ o The bandpass filter circuit has now been modified to te circuit above. If the angular frequency, ω = 2.86 10 6, find the equivalent impedance. CPEN 206 Lecture 3 2015_2016 29
Hybrid Example 5 10V V 1 V 2kΩ 2 V 2kΩ 3 o Find V 1, V 2, and V 3 CPEN 206 Lecture 3 2015_2016 30
Hybrid Solution 5 10V V 1 V 2kΩ 2 V 2kΩ 3 o Find an equivalent resistance for the network with V 1 across it, then find V 1 using a voltage divider. CPEN 206 Lecture 3 2015_2016 31
Hybrid Solution 5 10V V 1 V = 10V 1 = 5V CPEN 206 Lecture 3 2015_2016 32
Hybrid Solution 5 10V 5V 2kΩ V 2 2kΩ V 3 o Find an equivalent resistance for the network with V 2 across it, then find V 2. CPEN 206 Lecture 3 2015_2016 33
Hybrid Solution 5 10V 5V 2kΩ V 2 V 5V 2 = = 2.5V CPEN 206 Lecture 3 2015_2016 34
Hybrid Solution 5 10V 5V 2.5V 2kΩ 2kΩ V 3 V 2.5V 3 = = 1.25V CPEN 206 Lecture 3 2015_2016 35
Hybrid Example 6 100Ω 0.1Ω 70.4mH 10V 0 100µF V out o The circuit diagram above is a sample notch filter circuit network for processing signal. Find the output voltage of the network, V out if the angular frequency ω = 1500 and the input voltage is 10V. CPEN 206 Lecture 3 2015_2016 36
Hybrid Solution 6 100Ω 0.1Ω j106ω 10V 0 j6.67ω V out o Find the equivalent impedance of the resistor, inductor, and capacitor. CPEN 206 Lecture 3 2015_2016 37
Hybrid Solution 6 100Ω 7.12Ω 89.99 10V 0 V out o Combine resistor and impedance. CPEN 206 Lecture 3 2015_2016 38
Hybrid Solution 6 100.3Ω 4.07 10V 0 V out CPEN 206 Lecture 3 2015_2016 39
Hybrid Solution 6 V out 0 = 10V 0 100.3Ω 4.07 0 V out = 9.09V 0. 37 CPEN 206 Lecture 3 2015_2016 40