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Answering questions

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Lab 5 Worksheet Note: Lab 5 is generally the most challenging exercise in ENGR 2105. Please

watch the videos and read Lab 5 carefully at least twice and then take your time on

the exercises below to make sure that you understand the theoretical material.

1. In what quadrant of the complex plane are these numbers located?

−12+j7 __________________ −10−j50 __________________

8−j2 __________________ 1+j100 __________________

2. Use the complex conjugate to convert the expressions to a real term and an

imaginary term.

26 (6 − 𝑗4)⁄ __________________ (8 − 𝑗8) (2 + 𝑗2)⁄ __________________

3. Inductor and capacitor impedances are given as: 𝑍𝐿 = 𝑗𝜔𝐿 and 𝑍𝐶 =

1 𝑗𝜔𝐶⁄ . Assume you have a 10μF capacitor and a 10mH inductor. Calculate the impedance of these components at the following frequencies and list in

the space provided:

1 MHz (1,000,000 Hz): 𝑗𝜔𝐿 = _______________Ω 1 𝑗𝜔𝐶⁄ = _________ Ω

50KHz (50,000 Hz): 𝑗𝜔𝐿 = ______________ Ω 1 𝑗𝜔𝐶⁄ = ___________ Ω

0Hz: 𝑗𝜔𝐿 = _______________ Ω 1 𝑗𝜔𝐶⁄ = _______________ Ω

4. Different items in the time domain transform in different ways to the ω

domain:

Element Time Domain ω Domain Transform

Applied Sinusoidal

AC Voltage

𝑽𝒑 𝐜𝐨𝐬(𝝎𝒕)

(Volts)

Vp (Volts)

Series Current 𝑰𝒑 𝐜𝐨𝐬(𝝎𝒕 + 𝜽)

(Amps)

𝑰𝒑

(Amps)

Resistance R

(Ohms)

R

(Ohms)

Inductance L

(Henry’s) 𝒋𝝎𝑳

(Ohms)

Capacitance C

(Farads) 𝟏 𝒋𝝎𝑪⁄ (Ohms)

© N. B. Dodge 01/12

Given a circuit with a Time domain values:

𝑣(𝑡) = 10 cos(1000𝑡)

𝑅 = 100 Ω

L = 10 mH

C = 10 μF

Calculate the values in the ω domain for:

Applied Voltage __________Volts

Resistance __________Ω

Impedance from Inductor __________ Ω

Impedance from Capacitor __________ Ω

5. After transforming voltage and circuit to the ω domain, find the current by

dividing voltage by impedance. This usually results in a complex number. To

convert back to the time-domain, which is the answer sought, do four things:

• Rationalize using complex conjugate; the result is an X ± jY representation.

• Convert complex number to polar-coordinates:

• The radial distance is the peak current. The angle is the phase angle.

6. Based on the procedure in 5 above, convert the following ω domain currents

back into the time domain (assume ω = 1000):

I = 10+j10 _____________ I = −8+j4 _____________

7. If 𝑣(𝑡) = 10 cos(1000𝑡), R = 100 Ω, C = 100 μF , determine the expression for i(t) .

© N. B. Dodge 01/12

ENGR 2105 – Inductors and Capacitors in AC Circuits and Phase Relationships

1. Introduction and Goal: Capacitors and inductors in AC circuits are studied. Impedance and phase relationships of AC voltage and current are

defined. Frequency-dependence of inductor and capacitor impedance is

introduced. Phase relationships of AC voltage and current are defined.

2. Equipment List:

• Multisim

• Scientific Calculator

3. Experimental Theory: Capacitors and inductors change the voltage- current relationship in AC circuits. Since most single-frequency AC circuits

have a sinusoidal voltage and current, exercises in Experiment 5 use

sinusoidal AC voltages. Note that in an RLC AC, current frequency will be

identical to the voltage, although the current waveform will be different.

“Imaginary” Numbers, the Complex Plane, and Transforms:

3.1.1 Definition of j: As √−1 is not a real number, Electrical

Engineers define 𝑗 = +√−1. Physicists and

Mathematicians use 𝑖 = −√−1 for this same purpose, so 𝑗 = −𝑖, but that will not affect our theory.

3.1.2 Electrical Engineering problem solutions often include imaginary numbers. It is useful to consider real and

imaginary numbers as existing in a two dimensional space,

one axis of which is a real-number axis, and the other of

which is the “imaginary” axis.

3.1.3 In the complex plane (Figure 1), the horizontal axis is the real axis, and the vertical axis is the y axis. Real numbers (–7 , 10) lie on the x-

axis, imaginary numbers (–3j, j42) lie on the y-axis. Complex

numbers lie off axis. For example, 2 + j6 would lie in the first

quadrant, and (–43 – j17) would lie in the third quadrant.

© N. B. Dodge 01/12

3.1.4 Transforms: Transforms allow moving a problem from a coordinate system or domain where it is difficult to solve to

one where it is easier to solve (Figure 2).

3.1.5 Simpler equations in the transform domain make the problem easier to solve than in the original domain. We can

transfer sinusoidal, single-frequency AC circuit problems to

a domain where we can use algebra to solve them rather than

calculus. Solving problems in algebra is almost always

easier than calculus!

3.1.6 The catch: We need transforms (formulas) to the new

domain. Then “inverse transforms” are required to return the

solution to the time domain, where it is useful.

3.2 A New Domain: In the phasor or frequency (ω) domain, sinusoidal AC circuit problems are easier to solve. Note: 𝛚 = 𝟐𝛑𝐟, where f is the frequency in Hertz. Recall that Hertz has units of 1/second, or “per

second”. Thus, ω is in radians/sec.

3.2.1 Transforms: Skipping the derivation (you will do the derivation if you take ENGR 2305), we simply list

frequency domain transforms. Note that AC voltage is

© N. B. Dodge 01/12

usually expressed as 𝑣(𝑡) = 𝑉𝑝cos(ωt), where Vp is the peak

voltage.

3.3 Transform from the Time Domain to the Frequency Domain: Some circuit elements have a different representation in the frequency, or ω,

domain.

Element Time Domain ω Domain Transform

Sinusoidal AC Voltage 𝑉𝑝cos(ωt) Vp (Volts)

Resistance R (Ohms) R (Ohms)

Inductance L (Henry’s) 𝑗𝜔𝐿 (Ohms) Capacitance C (Farads) 1 𝑗𝜔𝐶⁄ (Ohms)

3.4 Comments: 3.4.1 Resistance: There is no transform for Resistors. We use

the resistance value R (Ohms) in both domains.

3.4.2 Inductance: In the time domain, we use the value of the Inductor, L (Henry’s). This value transforms to 𝑗𝜔𝐿 in the ω domain. 𝑗𝜔𝐿 has the same units as Resistance and represents the amount the Inductor opposes current.

3.4.3 Capacitance: In the time domain, we use the value of the Capacitor, C (Farads). This value transforms to 1 𝑗𝜔𝐶⁄ in the ω domain. 1 𝑗𝜔𝐶⁄ has the same units as Resistance and represents the amount the Capacitor opposes current.

3.4.4 Voltage: Voltage in the ω domain is just the peak voltage, Vp. There is no frequency information in the voltage

transform.

3.5 Solving For Currents in the ω Domain: In most Electrical Engineering problems, we know voltages and component values: Resistance (R),

Inductance (L), and Capacitance (C). We generally solve for circuit

currents, which is easier in the frequency domain. Note: V = IR in the

time domain; V = IZ in the ω domain, where Z is the circuit

impedance. It is the sum of all the impedance contributions from

resistors, capacitors and inductors, which all have the unit of Ohms.

𝒁 = 𝑹 + 𝒋𝛚𝐋 + 𝟏 𝒋𝛚𝐂⁄ . In both domains, voltage is still in Volts and current is still in

Amperes.

3.5.1 Resistor AC circuit solution in the ω domain: In Figure 3, we identify time domain voltage (𝑣(𝑡) = 𝑉𝑝𝑐𝑜𝑠(𝜔𝑡) =

10cos⁡(1000𝑡). From this, we obtain the ω domain voltage

© N. B. Dodge 01/12

transform: Vp = 10 Volts and ω = 1000 radians/second.

Resistance = 100 Ω. Since V=I Z in the ω domain, then

𝐼 = 𝑉𝑝 𝑍⁄ = 𝑉𝑝 𝑅⁄ = 10 100⁄ = 0.1⁡𝐴𝑚𝑝𝑠

This answer is converted to the time domain later in the lab.

3.5.2 Inductor AC circuit (Figure 4): 𝑉 = 10cos(1000𝑡), so ω =

1000 rad/s. In the ω domain, 10 mH transforms to jωL = j(1000)(10 ∗ 10−3) = 𝑗10 Ohms. Vp = 10 Volts. So

𝐼 = 𝑉𝑝 𝑍⁄ = 𝑉𝑝 𝑗ωL⁄ = 10 𝑗10⁄ = −𝑗1⁡𝐴𝑚𝑝𝑠

(Time Domain answer below.)

3.5.3 Capacitor AC circuit (Figure 5): ⁡𝑉 = 10cos(1000𝑡), so ω

= 1000 rad/s. In the ω domain, 100 μF transforms to

1 𝑗⁄ ωC = 1 j(1000)(100 ∗ 10−6)⁄ = −𝑗10 Ohms in the ω domain. Vp = 10 Volts. So

𝐼 = 𝑉𝑝 𝑍⁄ = 𝑉𝑝 𝑗ωL⁄ = 10 −𝑗10⁄ = 𝑗1⁡𝐴𝑚𝑝𝑠

(Time Domain answer below.)

3.5.4 RL Circuit: Resistor and Inductor: In Fig.6, R and L

transform to 10 Ω and 𝑗ωL = j(1000)(10 ∗ 10−3)=j10 Ω. Then

© N. B. Dodge 01/12

𝐼 = 𝑉𝑝 𝑍⁄ = 10 (10 + 𝑗10)⁄ =

10(10 − 𝑗10) (100 + 100)⁄ = 0.5 − 𝑗0.5 Amps

3.6 Inverse Transforms: To make solutions from the ω domain useful, we must do the reverse (or inverse) transform to the time domain.

3.6.1 The answers above are in Cartesian coordinates (X ± jY). These X ± jY results would be more useful in polar

coordinates. We want to find r (r is actually the peak

current, Ip) and θ.

𝑟 = 𝐼𝑝 =

√(|”𝑅𝑒𝑎𝑙”⁡𝐶𝑢𝑟𝑟𝑒𝑛𝑡|)2 + (|”𝐼𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦”⁡𝐶𝑢𝑟𝑟𝑒𝑛𝑡|)2

𝜃 = tan−1 (|”𝐼𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦⁡𝐶𝑢𝑟𝑟𝑒𝑛𝑡”|) |”𝑅𝑒𝑎𝑙”⁡𝐶𝑢𝑟𝑟𝑒𝑛𝑡|⁄

3.6.2 The time domain current is then: 𝐼(𝑡) = 𝐼𝑝cos⁡(ωt + θ)

3.6.3 Resistor Circuit: The current had only a Real component. I = 0.1 Amps.

𝑟 = 𝐼𝑝 = √(0.1) 2 + (0)2 = 0.1⁡𝐴𝑚𝑝𝑠

𝜃 = tan−1 0 0.1⁄ = 0⁡𝐷𝑒𝑔𝒓𝒆𝒆𝒔 𝑰(𝒕) = 𝑰𝒑 𝐜𝐨𝐬(𝛚𝐭 + 𝛉) = 𝟎.𝟏𝐜𝐨𝐬(𝟏𝟎𝟎𝟎𝒕)

The voltage in the time domain is 𝑉 = 10cos(1000𝑡). Comparing this to the current, we see that V and I have different amplitudes, but they will rise

and fall “in phase” with each other, as shown in Figure 7.

© N. B. Dodge 01/12

Fi

3.6.4 Inductor Circuit: The current had only an Imaginary component. I = -j 1 Amps.

𝑟 = 𝐼𝑝 = √(0) 2 + (1)2 = 1⁡𝐴𝑚𝑝𝑠

𝜃 = tan−1 −1 0⁄ = −90⁡𝐷𝑒𝑔𝑟𝑒𝑒𝑠 = −𝜋 2⁄

𝑰(𝒕) = 𝑰𝒑𝐜𝐨𝐬⁡(𝛚𝐭 + 𝛉) = 𝟏𝐜𝐨𝐬(𝟏𝟎𝟎𝟎𝒕 − 𝝅 𝟐⁄ )⁡𝑨𝒎𝒑𝒔

The result is sinusoidal current with a peak value of 1

ampere, with an associated phase angle. That is, it oscillates

at the same radian frequency of 1000 rad/s, but is not in

lock-step with the voltage. Its oscillation is 90 degrees

behind the voltage, as shown in the graph below (Figure 8).

This “phase angle” is constant. Current is always exactly 90

degrees behind the voltage, a significant characteristic of

inductors in sinusoidal AC circuits

3.6.5 Capacitor Circuit: The current had only an Imaginary

component. I = j 1 Amps.

𝑟 = √(0)2 + (1)2 = 1⁡𝐴𝑚𝑝𝑠 𝜃 = tan−1 1 0⁄ = +90⁡𝐷𝑒𝑔𝑟𝑒𝑒𝑠 = + 𝜋 2⁄

𝑰(𝒕) = 𝑰𝒑𝐜𝐨𝐬⁡(𝛚𝐭 + 𝛉) = 𝟏𝐜𝐨𝐬(𝟏𝟎𝟎𝟎𝒕 + 𝝅 𝟐⁄ ) Amps

© N. B. Dodge 01/12

The result is sinusoidal current with a peak value of 1

ampere, with an associated phase angle. That is, it oscillates

at the same radian frequency of 1000 rad/sec, but is not in

lock-step with the voltage. Its oscillation is 90 degrees ahead

of the voltage, as shown in the graph below (Figure 9). This

“phase angle” is constant. Current is always exactly 90

degrees ahead of the voltage, a significant characteristic of

Capacitors in sinusoidal AC circuits

3.6.6 RL Circuit: The current had both a Real and an Imaginary

component. I = 0.5 – j 0.5 Amps.

𝑟 = √(0.5)2 + (0.5)2 = 0.707⁡𝐴𝑚𝑝𝑠 𝜃 = tan−1 −0.5 0.5⁄ = −45⁡𝐷𝑒𝑔𝑟𝑒𝑒𝑠 = − 𝜋 4⁄

𝑰(𝒕) = 𝑰𝒑𝐜𝐨𝐬⁡(𝛚𝐭 + 𝛉) = (𝟎.𝟕𝟎𝟕)𝐜𝐨𝐬(𝟏𝟎𝟎𝟎𝒕 − 𝝅 𝟒⁄ )⁡𝑨𝒎𝒑𝒔

The result is sinusoidal current with a peak value of 0.707

Amperes, with an associated phase angle. That is, it

oscillates at the same radian frequency of 1000 rad/s, but is

not in lock-step with the voltage. Its oscillation is 45 degrees

behind the voltage. This “phase angle” is constant. Current

is always exactly 45 degrees behind the voltage.

4. Pre-Work: Prior to lab, review the experimental theory and experimental procedure and complete the Worksheet.

5. Experimental Procedure: 5.1 V-I We first study an RL circuit. For the following, use 𝜔 =

1000 𝑟𝑎𝑑

𝑠 ⁡

5.1.1 Construct a series RL circuit as in Fig. 6, using a 10 mH inductor and 16 Ω resistor. Recall that the frequency can be

calculated as 𝑓 = 𝜔

2𝜋 . Connect ground between the inductor

and the AC Voltage source. Connect a combined voltage

and current probe between the AC source and the resistor

© N. B. Dodge 01/12

(This setup will simultaneously measure the output of the

AC Voltage Source and the total current in the circuit).

5.1.2 Using time cursors, measure time difference, ∆𝑡, between Ip (on the “Current” Curve) and Vp (on the “Voltage” curve).

This ∆𝑡 will be used to determine the phase angle. 5.1.3 Take a screen shot of your Multisim setup for this

measurement and include it in your Lab Deliverables.

5.2 Examining V-I Relationship in an AC RC Circuit: Replace the inductor with a 10 μF capacitor to create an RC circuit. Leave the 16

Ω resistor in place and continue using 𝜔 = 1000 𝑟𝑎𝑑

𝑠 .

5.2.1 Using the cursors, measure “∆𝑡” between current and voltage peaks.

5.2.2 Take a screen shot of your Multisim setup for this measurement and include it in your Lab Report.

5.2.3 ∆𝑡 is used to calculate the phase angle. We first find the ratio of ∆𝑡 to the total amount of time required to complete one period, T. Where 𝑇 = 1 𝑓⁄ , and 𝑓 = 1000 Hz for this experiment.

𝑅𝑎𝑡𝑖𝑜 = ∆𝑡 𝑇⁄ = ∆𝑡 (1 𝑓⁄ )⁄

5.2.4 This ratio must also represent the ratio of the phase shift, θ, to 360 degrees:

𝑅𝑎𝑡𝑖𝑜 = 𝜃 360𝑜⁄ 5.2.5 Equating the two terms, ∆𝑡 𝑇⁄ = 𝜃 360𝑜⁄ . Rearranging, we

can solve for the phase shift:

𝑃ℎ𝑎𝑠𝑒⁡𝑆ℎ𝑖𝑓𝑡⁡𝜃 = (∆𝑡 𝑇⁄ ) ∗ 360𝑜

6. Cleanup: Return parts and cables to their respective homes. Make sure that your work area is clean.

7. Writing the Lab Report: For your lab deliverables, do the following: 7.1 Since 𝑖(𝑡) = 𝐼𝑝cos(𝜔𝑡 + 𝜃), construct expressions for i(t) in the RL

and RC circuits using the given circuit values. Do this by

transforming to the 𝜔-domain and calculating 𝐼𝑝 and 𝜃, and then

transform back to the time domain.

7.2 From your measurements, write an expression for i(t) in each case. 7.3 Compare the i(t) expressions developed in 7.1 and 7.2. Discuss any

discrepancies.

Experiment #5 Data Sheet

1. RL Circuit measurements: 2.1.Measured peak (Vp) voltage of the Voltage Source: __________

2.2.Peak current (A): __________

2.3.Time delta (μsec) between current and voltage peaks: __________

2.4.Time-domain expression for i(t), based on measures above: __________

2.5.Calculated expression for i(t), based on measured R and L: __________

2.6.List your ideas for discrepancies, if any: _________________________

2. RC Circuit measurements:

2.1 Measured peak voltage of the Voltage Source: __________

2.2 Peak current (A): __________

2.3 Time delta (μsec) between current and voltage peaks: __________

2.4 Time-domain expression for i(t), based on measures above: __________

2.5 Calculated expression for i(t), based on measured R and C: __________

2.6 List your ideas for discrepancies, if any: _________________________

_____________________________________________________________

Brief 5 AC RL and RC Circuits Electrical Circuits Lab I (ENGR 2105)

Dr. Kory Goldammer

Review of Complex Numbers and Transforms

Transforms

The Polar Coordinates / Rectangular Coordinates Transform

The Complex Plane

We can use complex numbers to solve for the phase shift in AC Circuits

Instead of (x,y) coordinates, we define a point in the Complex plane by (real, imaginary) coordinates

Real numbers are on the horizontal axis

Imaginary numbers are on the vertical axis

The Complex Plane (cont.)

Imaginary numbers are multiplied by j

By definition,

(Mathematicians use i instead of j, but that would confuse us since i stands for current in this class)

Imaginary Plane: Rectangular Coordinates

We can identify any point in the 2D plane using (real, imaginary) coordinates

Complex Plane Using Rectangular Coordinates

Imaginary Plane: Transform to Polar Coordinates

We can identify any point in the complex plane using (r,) coordinates.

The arrow is called a Phasor.

r is the length of the Phasor, and is the angle between the Positive Real Axis and the Phasor

is the Phase Angle we want to calculate

Complex Plane Using Rectangular Coordinates

r

Imaginary Plane: Transform to Polar Coordinates

Complex Plane Using Rectangular Coordinates

(we will discuss the meaning of r later)

Use either the sin or cos term to find :

But we need in radians:

r

=7.07

Complex Math

For addition or subtraction, add or subtract the real and j terms separately.

(3 + j4) + (2 – j2) = 5 + j2

To multiply or divide a j term by a real number, multiply or divide the numbers. The answer is still a j term.

5 * j6 = j30

-2 * j3 = -j6

j10 / 2 = j5

Complex Math (cont. 1)

To divide a j term by a j term, divide the j coefficients to produce a real number; the j factors cancel.

j10 / j2 = 5

-j6 / j3 = -2

To multiply complex numbers, follow the rules of algebra, noting that j2 = -1

Complex Math (cont. 2)

To divide by a complex number: Can’t be done!

The denominator must first be converted to a Real number!

Complex Conjugation

Converting the denominator to a real number without any j term is called rationalization.

To rationalize the denominator, we need to multiply the numerator and denominator by the complex conjugate

Complex Number Complex Conjugate

5 + j3 5 – j3

–5 + j3 –5 – j3

5 – j3 5 + j3

–5 – j3 –5 + j3

Complex Math (cont. 2)

Multiply the original equation by the complex conjugate divided by itself (again, j2 = -1):

Phase Shift Time Domain – ω Domain Transforms

Transforming from the Time (Real World) Domain to the (Problem Solving Domain

Note that in the ω Domain, Resistance, Inductance and Capacitance all of units of Ohms!

ElementTime Domainω Domain Transform
Applied Sinusoidal AC Voltage(Volts) (ω=2πf)Vp (Volts)
Series Current(Amps) (ω=2πf)(Amps)
ResistanceR (Ohms)R (Ohms)
InductanceL (Henry’s)(Ohms) (ω=2πf)
CapacitanceC (Farads)(Ohms) (ω=2πf)

Solving For Current

Impedance – Ohms

So far, we’ve talked about resistance. Ohm’s law tells us:

Resitance limits, or “impedes”, the flow of current.

The more Ohms in the circuit, the less current.

Impedance (Z): The number of Ohms in a circuit.

Resistance has units of Ohms, but so do Capacitance and Inductance in the ω Domain!

Impedance – Ohms (cont. 1)

The impedance of the Capacitor:

When the frequency is small, ZC is large.

When the frequency is large, ZC is small

The impedance of the Inductor:

When the frequency is large, ZL is large.

When the frequency is small, ZL is small

Impedance – Ohms (cont. 2)

To find the total impedance of a series circuit:

ElementTime Domainω Domain Transform
Applied Sinusoidal AC Voltage(Volts) (ω=2πf)Vp (Volts)
Series Current(Amps) (ω=2πf)(Amps)
ResistanceR (Ohms)R (Ohms)
InductanceL (Henry’s)(Ohms) (ω=2πf)
CapacitanceC (Farads)(Ohms) (ω=2πf)

Ohm’s Law Still Works

Ohm’s law is well-named. It applies to anything that has units of Ohms!

Instead of , it becomes:

Or,

Ohms law works in both the time- and ω-domains!

Strategy

Solving in the time domain is difficult!

Transform v(t) to the ω domain

Calculate Z in the ω domain

Solve for Current in the ω domain:

Transform Current back to the time domain

Calculating current in an R (Resistor Only) Circuit – Nothing Changes!

Time Domain

ω Domain

Vp

Phase Angle for an R Circuit

Time Domain

0.1

Interpretation: The current in the circuit will be in phase with the applied voltage.

+j

+

I = 0.1 Amps

θ = 0o

Calculating current in an L (Inductor Only) Circuit

Time Domain

ω Domain

Vp

Phase Angle for an L Circuit

Time Domain

1

Interpretation: The current in the circuit will Lag the voltage by 90o (i.e. Current reaches it’s peak at a later time)

+j

+

I = -j1 Amps

Calculating current in a C (Capacitor Only) Circuit

Time Domain

ω Domain = 10 Volts

Vp

Phase Angle for a C Circuit

Time Domain

1

Interpretation: The current in the circuit will Lead the voltage by 90o

+j

+

I = j1 Amps

Calculating current in an RL Series Circuit

Time Domain

ω Domain =10Volts

Vp

Calculating Current for RL Circuit

Rectangular: Amps

Polar:

But we need in radians:

+j

+

θ

)

I =0.5 – j0.5 Amps

Calculating Current for RL Circuit – Time Domain

The Current is out of phase with the applied voltage by 45o

Transform Theta from Radians to Time

In the previous example, we said the two waveforms were out of phase by

How much is the phase shift in terms of time?

We know in 1 Period, the wave goes through , or 360o. So in this case, the shift is:

We know 1 period takes

3) The phase shift is of a period, so

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