Series-parallel circuits
Series-parallel circuits
[Q1], [Q5] Series-parallel circuit are combo of series and parallel circuits--current flows in some components sequentially (series), some split to circuit's other parts (parallel).
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[5] How to solve a combo circuit?
Reduce the circuit to a series circuit by identifying and adding loads in parallel and series together.
Calculate RT (total resistance).
Calculate IT (total current).
Expand and solve for remaining values.
Rules
Rules
Removing loads:
Removed resistance increase IT.
Removed/added resistance won't affect VT/voltage source.
Removed resistance increase each resistors' voltage drop.
Removed resistance increase PT (total power) due to higher current.
Added resistance increase individual voltage values.
Added resistance reduce IT.
Added resistance decrease PT (total power) due to lower current.
Loads in series:
Current is the same
Voltage drops add to source/VT
Resistances sum is RT
VT the same across loads in a loop.
Loads in parallel:
Voltage values per parallel branches are the same.
RT is always less than individual resistance.
Current diminish as it ‘splits’ per branches.
E.g. To find all voltages and currents of this circuit:
Since all currents are needed, reduce the circuit to find the total resistance.
Often, start the furthest away from the source and move to it.
Current flows and polarity
Current flows and polarity
Note: Arrows are currents' paths
IT value exiting the battery's positive terminal = same current value returns/enters in its negative terminal.
Current enters the 220 Ω resistor, so its left is positive and right side is negative.
IT splits at R1, R2, R3 and R5 nodes.
Current value entering a node = current value leaving it
All different currents in this circuit.
Example 1
Example 1
Find RT:
Start furthest away from the source.
Combine what's in parallel/series into 1 resistor step by step (simplify the circuit).
Parallel resistors: R5 and R4, R4 and R6.
Between R3 and R5: R3||R5 = 100||2000 = 95.24 Ω
Between R4 and R6: R4||R6 = 680||1200 = 434.04 Ω
R3||R5 and R4||R6 are now in series, add them:
434.04 + 95.24 = 529.28
330||529.28 = 203.27
RT = 203.27 + 220 = 423.27 Ω
Now redraw the circuit.
2. These 2 resistors are in series. 95.2 Ω resistor's current also goes into 434 Ω resistor.
R3||R5 + R4||R6 = 95.2 + 434 = 529.2 Ω
Combine the 2 resistors and redraw the circuit.
3. These 2 are in paralell: 330||529.3 = 203.3 Ω
Combine and redraw.
4. These 2 are in series: 220 + 203.3 = 423.3 Ω
Combine and redraw.
Calculating IT
Calculating IT
5. Calculate the total current: IT = VT/RT = 9/423.3 = 0.0212615167 A = 21.26 mA
IT in the equivalent RT (final version of the circuit) is the same current as going in this circuit (2 resistors in series = total resistance).
Find the voltage drop per resistor.
Convert mA to A: 21.3 mA = 0.0213 A
R1V = 0.0213 x 220 = 4.7 V
R2(R3||R5 + R4||R6)V = 0.0213 x 203.3 = 4.3 V
203.3Ω resistor is made of 2 parallel resistors as shown.
4.3 V across the equivalent 203.3 Ω resistor is the same as across the 330 Ω and the 529.3 Ω resistor.
Find the branches' each resistors' currents:
Green arrow: I = V/R = 4.3/529.3 = 0.00812393728 A = 8.1 mA
Red arrow: 4.3/330 = 0.013030303 A = 13 mA
529.3 Ω resistor = 2 resistances values in series.
All resistors have 8.1 mA flowing through.
Example 2
Example 2
Calculate
RT
IT
Voltage per branch
Each load's current and powers
Voltage drop across per load
Recall: Voltage drop across each load is voltage reduction as current goes through loads (resistor/light bulbs) due to load's resistance.
Power dissipated across per load
Reimagine the circuit in a basic parallel circuit's structure.
Calculating RT
Calculating RT
Resistors in parallel:
R2+R3 = 60+25 = 85
R4+R5 = 50+75 = 125
R6+R7 = 40+100 = 140
Redraw:
Resistors in parallel:
140||125||85 = 37.16
Redraw:
Resistors in series:
RT = R1+37.16 = 100+37.16 = 137.16 Ω
Summary of calculating RT and simplifying.
Calculating IT
Calculating IT
IT (and IR1) = VT/RT = 20/137.16 = 0.145815106 A = 145.82 mA
Calculating current per branch
Calculating current per branch
VR1 = ITxR1 = 0.145815106 x 100 = 14.58 V
Voltage per branch: VT-VR1 = 20-14.58 = 5.42 V--Each branches have 5.42 V:
VR2+VR3 = 5.42 V
VR4+VR5 = 5.42 V
VR6+VR7 = 5.42 V
Calculating current per load
Calculating current per load
Divide each resistor by the voltage per branch
(5.42V ÷ branch resistance):
IR2 = IR3 = 5.42/(IR2 + IR3) = 5.42/(60+25) = 0.0637647059 mA = 63.76 A
IR4 = IR5 = 5.42/(IR4 + IR5) = 5.42(75+50) = 0.04336 mA = 43.36 A
IR6 = IR7 = 5.42/(IR6 + IR7) = 5.42/(40+100) = 0.0387142857 mA = 38.71 A
Calculating voltage per load
Calculating voltage per load
Divide each resistors by their own voltages:
VR2 = 0.0637647059 x 60 = 3.83 V
VR3 = 0.0637647059 x 25 = 1.59 V
VR4 = 0.04336 x 50 = 2.17 V
VR5 = 0.04336 x 75 = 3.25 V
VR6 = 0.0387142857 x 40 = 1.55 V
VR7 = 0.0387142857 x 100 = 3.87 V
Calculating power per load and PT
Calculating power per load and PT
PR1 = 14.58 x 0.145815106 = 2.12598425 W
PR2 = 0.0637647059 x 3.84 = 0.24 W
PR3 = 0.0637647059 x 1.59 = 0.10 W
PR4 = 0.04336 x 2.17 = 0.09 W
PR5 = 0.04336 x 3.25 = 0.14 W
PR6 = 0.0387142857 x 1.55 = 0.06 W
PR7 = 0.0387142857 x 3.87 = 0.15 W
PT = PR1+PR2+PR3+PR4+PR5+PR6+PR7 = 2.92 W
Example 3
Example 3
Calculate:
RT
IT
VT
Voltage drop per load
PT
Power per load
In parallels: R2+R3+R4 = 75||50||40 = 17.14 Ω
In series: RT = 200+17.14+240 = 457.14 Ω
IT = 120/457.14 = 0.262501641 A or 262.5 mA
VT = 0.262501641 x 457.14 = 120 V (confirmed)
PT = 0.262501641 x 120 = 31.5 V
Power drop per load:
PR1 = 16.54 W
PR2 = 4.5
PR3 = 4.5
PR4 = 4.5
PR5 = 0.262501641
Voltage drop per load:
VR1 = 63
VR2 = 4.5
VR3 = 4.5
VR4 = 4.5
VR5 = 52.5
All values
Example 4
Example 4
[1]
Calculate all values.
RT = 1.36 kΩ
IT = 22.0 mA
IR1 = 22.0 mA
IR2 = 13.11 mA
IR3 = 8.94 mA
VR1 = 10.34 V
VR2 = 19.66 V
VR3 = 19.66 V
PR1 = 227.5 mW
PR2 = 257.7 mW
PR3 = 175.8 mW
PT = 660 mW
Fig 1
"Textbook exercises" resistor network
"Textbook exercises" resistor network
Some exotic circuits with resistors like these (Fig 1) are mainly for math purposes in textbooks.
To calculate their IT:
Note: current value entering node = current value leaving node
current value splits = substracted
E.g. 1# Work from left to right, 6A enters the node and 2A leaves: 1st unknown resistor ("? A") = 6-4 = 2 A.
At the first node, the current reunite (2+4=6).
2nd "? A" 6-1.5 = 4.5 = A
E.g. 2# 1st "? A" = 3 A
2nd "? A" = 12 A
3rd "? A"= 4 A
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