Direct Current Circuits

• The content covers Direct Current Circuits

Objectives

• Determine the effective resistance for resistors in series
• Determine the effective resistance for resistors in parallel
• Determine voltage and current for each resistor in simple and complex circuits
• Use Kirchoff's laws to find currents and voltages in complex circuits

Electrical Circuit Symbols

• Electrical circuits have resistors grouped together
• Resistors are attached to an energy source, such as a battery
• Ground symbol is a horizontal line connected to three descending, progressively shorter horizontal lines.
• Battery symbol is a long vertical line and a short vertical line
• The resistor symbol is a jagged line or rectangle

Resistances in Series

• Resistors in series means there is a single path for the current
• The current (I) is the same for each resistor (R1, R2, and R3)
• The energy gained (E) is lost through the resistors (R1, R2, and R3,)
• The same is true for voltages

Series Connection Equation

• I = I1 = I2 = I3
• VT = V1 + V2 + V3

Equivalent Resistance Equation

• The equivalent resistance (Re) of resistors connected in series equals individual resistances sum
• VT = V1 + V2 + V3
• ITRe = I1R1+ I2R2 + I3R3
• Re = R1 + R2 + R3

Measuring Equivalent Resistance

• Find equivalent resistance Re, finding current (I) in the circuit
• Re = R1 + R2 + R3
• Re = 3 Ω + 2 Ω + 1 Ω = 6 Ω
• The current can be found using Ohm’s law V = IRe
• If the voltage is 12 then, I = V/Re = 12V / 6 Ω
• I = 2 A

Voltage Drops with Three Resistors

• The voltage drops across the three resistors totals the 12-V emf
• Re = 6 Ω
• I = 2 A
• The current I = 2 A is the same in each R
• V1 = IR1; V2 = IR2; V3 = IR3
• V1 = (2 A)(1 Ω) = 2 V
• V1 = (2 A)(2 Ω) = 4 V
• V1 = (2 A)(3 Ω) = 6 V
• V1 + V2 + V3 = VT
• 2 V + 4 V + 6 V = 12 V

EMF in Series

• The output direction from a source of emf is from the positive (+) side
• From point a to b the potential increases by E
• From point b to a, the potential decreases by E

Finding ΔV

• Find ΔV for path AB and then for path BA
• AB: ΔV = +9 V – 3 V = +6 V
• BA: ΔV = +3 V - 9 V = -6 V

Single Complete Circuit

• Consider the simple series circuit
• Path ABCD energy and voltage increase through the 15V source
• Voltage decreases through the 3V source down to 12V
• The net gain in potential is lost through the two resistors
• IR2 and IR4 means the sum is zero for the entire loop

Finding Current in a Simple Circuit

• Find current I in the circuit below
• ΣE = 18V-3 V = 15 V
• ΣR = 3 Ω + 2 Ω = 5 Ω
• Applying Ohm’s law:
• I = ΣE/ΣR = 15 V/5 Ω
• I = 3 A
• For a single loop circuit: I = ΣE/ΣR

Single Loop Circuits Summary

• Resistance Rule: Re = ΣR
• Current: I = ΣE/ΣR
• Voltage Rule: ΣE = ΣIR

Complex Circuits

• A complex circuit contains more than a single loop
• Complex circuits have different current paths
• At junctions m and n: I1 = I2 + I3 or I2 + I3 = I1

Junction Rule

• ΣI (enter) = ΣI (leaving)

Parallel Connections

• Resistors are connected in parallel when there is more than one path for current
• For Parallel Resistors: V2 = V4 = V6 =
• IT = I2 + I4 + I6
• For Series Resistors: I2 = I4 = I6 = IT
• V2 + V4 + V6 = VT

Parallel Equivalent Resistance

• VT = V1 = V2 =
• IT = I1 + I2 + I3
• Ohm’s law: I = V/R
• 1/Re = 1/R1 + 1/R2 + 1/R3
• The equivalent resistance for Parallel resistors

Finding Equivalent Resistance for Three Resistors

• Find the equivalent resistance Re for the three resistors below.
• 1/Re = Σ1/Ri
• 1/Re = 1/R1 + 1/R2 + 1/R3
• 1/Re = 1/2 Ω + 1/4 Ω + 1/6 Ω = 0.500+0.250+0.167
• 1/Re = 0.917; Re = 1/0.917 = 1.09 Ω
• Re= 1.09 Ω
• For parallel resistors, Re is less than the least Ri.

Solving for Total Current

• Assume a 12-V emf is connected to the circuit
• Find the total current leaving the source of emf
• VT = 12 V; Re = 1.09
• V1 = V2 = V3 = 12 V
• IT = I1 + I2 + I3
• Ie = VT/Re = 12 V/1.09 Ω
• Total current: IT = 11.0 A

Resistor Currents

• Show that the current leaving the source IT is the sum of the currents through the resistors R1, R2, and R3.
• IT = 11 A; Re = 1.09
• V1 = V2 = V3 = 12 V
• IT = I1 + I2 + I3
• I1= 12 V / 2 Ω = 6 A
• I2 = 12 V / 4 Ω = 3 A
• I3 = 12 V / 6 Ω = 2 A
• 6 A + 3 A + 2 A = 11 A

Shortcut Two Parallel Resistors

• The equivalent resistance Re for two parallel resistors is the product divided by the sum.
• 1/Re = 1/R1 + 1/R2
• Re= R1R2 / R1 + R2
• Re = (3Ω)(6Ω) / 3Ω+6Ω
• Re = 2 Ω

Activity 3: DC Circuits (Series)

• In the diagram, R1 = 5 Ω, R2 = 10 Ω, and R3 = 15 Ω
• The battery supplies an emf of c = 0.30 V.
• What is the equivalent resistance, Re?
• What is the current through each resistor?
• What is the voltage drop across each resistor?
• What is the power expended in each resistor?

Series and Parallel Combinations

• In complex circuits resistors are connected in both series and parallel ways
• Using series and reducing resistances will reduce the circuit
• Series and parallel resistances reduce the circuit into a one source of emf circuit
• Results in one equivalent resistance

Finding Resistance

• Find the equivalent resistance for the circuit, assuming VT = 12 V.
• R3.6 = (3Ω)(6Ω) / 3Ω+6Ω = 2Ω
• Re = 4 Ω + 2Ω
• Re = 6 Ω

Current

• Find the total current IT.
• Re = 6 Ω
• I = VT/Re = 12 V/6 Ω
• IT = 2.00 A

Equivalent Resistance

• Find the currents and the voltages across each resistor
• I4 = IT = 2 A
• V4 = (2 A)(4 Ω) = 8 V
• The remainder of the voltage: (12 V – 8 V = 4 V) drops across EACH of the parallel resistors.
• V3 = V6 = 4 V
• V3,6 = I3,6R3,6 = (2 A)(2 Ω)

Resistors Summary

• Find the currents and voltages across each resistor.
• V4 = 8 V
• V6 = V3 = 4 V
• I3 = V3/R3 = 4 V/3Ω = 1.33 A
• I6 = V6/R6 = 4 V/6Ω = 0.667 A
• The junction rule is satisfied:
• ΣI (enter) = ΣI (leaving)
• IT = I4 = I3 + I6

Laws for DC Circuits

•  Kirchoff’s first law: The sum of the currents entering a junction equals the sum of the currents leaving
• Junction Rule: ΣI (enter) = ΣI (leaving)
•  Kirchoff’s second law: The sum of the emf’s around any closed loop must equal the sum of the IR drops around that same loop;
• Voltage Rule: ΣE = ΣIR

Emf’s Sign Conventions

• When applying Kirchoff’s laws there should be assumption a consistent and positive tracing direction
• When applying the voltage rule, consider emf’s positive if normal output direction of the emf is with the assumed tracing direction.
• Tracing from A to B consider emf direction positive
• Tracing from B to A consider emf direction negative

IR Drops in Circuits

• Consider voltage rule, and IR drops are positive if the assumed current direction is with the assumed tracing direction.
• If tracing from A to B, this IR drop is positive.
• If tracing from B to A, this IR drop is negative

Analyzing a Circuit

• Assume possible consistent flow of currents
• Indicate positive output directions for emf’s
• Indicate consistent tracing direction (clockwise)
• Junction Rule: I2 = I1 + I3
• Voltage Rule: ΣE = ΣIR
• E1 + E2 = I1R1 + I2R2

Kirchhoff’s Laws Examples

• Kirchoff’s Laws: Loop II
• Assume counterclockwise positive tracing
• Voltage Rule: ΣE = ΣIR
• E2 + E3 = I2R2 + I3R3
• If traced clockwise: - E2 - E3 = -I2R2 - I3R3

Finding Independant Equations

• Find equations for a loop
• Voltage Rule: ΣE = ΣIR
• E3 – E1 = -I1R1 + I3R3
• If traced clockwise: E1 - E3 = I1R1 - I3R3
•  four independent equations from Kirchoff’s laws:
• I2 = I1 + I3
• E1 + E2 = I1R1 + I2R2
• E2 + E3 = I2R2 + I3R3
• E3 - E1 = -I1R1 + I3R3

Kirchhoff with Circuits

• Use Kirchoff’s laws to find the currents in the circuit
• Junction Rule: I2 + I3 = I1
• Tracing clockwise
• obtains Voltage Rule: ΣE = ΣIR
• 12 V = (5 Ω)I1 + (10 Ω)I2
• Recalling that V/Ω = A,
• 5I1 + 10I2 = 12 A

Simplfying with Ohm's law

• Consider Loop II is traced clockwise
• Voltage Rule: ΣE = ΣIR
• 6 V = (20 Ω)I3 - (10 Ω)I2
• Simplify by dividing by 2
• Recognize V/Ω = A,
• 10I3 - 5I2 = 3 A

Solving for Currents

• The resultant equations, are solved for.
• I2 + I3 = I1
• 5I1 + 10I2 = 12 A
• 10I3 - 5I2 = 3 A
• Substitute Eq.(1) for I1 in 5(I2 + I3) + 10I2 = 12 A
• results in 5I3 + 15I2 = 12 A
• Multiple equation by scaling with Ohm's Law, and then subtracting and dividing scales the results

Current Results

• The resultant equations, are solved over the real field
• 30I3 - 15I2 = 9 A and 5I3 + 15I2 = 12
• Putting I3 = 0.6 A in 10(0.6 A) = 3 A
• results in I2= 0.600 A
• the solve from the substitution results in I1= 1.20 A

Formulas Summary

• Rules for a simple, single loop circuit containing a source of emf and resistors are summarized with:
• Resistance Rule: Re = ΣR
• Current : I = ΣE/ΣR
• Voltage Rule: ΣE = ΣIR

Resistors Connection

• For resistors connected in series:
• I = I1 = I2 = I3
• VT = V1 + V2 + V3
• Re = R1 + R2 + R3
• Re = ΣR

Parallel Conventions

• For resistors connected in parallel:
• V = V1 = V2 = V3
• IT = I1 + I2 + I3
• 1/Re = Σ1/Ri
• ReR1R2
• Re= / R1 + R2

Laws Sumarized

• Kirchhoff’s first law: The sum of the currents entering a junction is equal to the sum of the currents leaving that junction
• Junction Rule: ΣI (enter) = ΣI (leaving)
• Kirchhoff’s second law: The sum of the emf’s around any closed loop must equal the sum of the IR drops around that same loop.
• Voltage Rule: ΣE = ΣIR