# Electrical Engineering Mathematics Assignment (Math Problem Sample)

This is an Engineering Mathematics Problems solving Assignment recently done for an online client for their submittal for an electrical engineering coursework. Different numerical problems of basic electrical concepts have been solved like current and voltage waveforms, power signals, RL circuits and harmonics analysis.

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ELECTRICAL ENGINEERING MATHEMATICS ASSIGNMENT

MTH109-MOD3 CT

Date:28-02-2019

Submitted BY:

XYZ

Task-1

* A current waveform may be described by

is=13cos2πft-π4 [A]

Where frequency f=1Hz, and t represents time.

Make time t the subject of the formula and hence determine the first point in time when the current waveform has the magnitude of +10A.

Solution:

As the frequency of the above expression is 1Hz, the current will be expressed as under:

is=13cos2πt-π4

So, to calculate the value of time t at the specified current +10A, we can put this in place of current and evaluate for t as under:

10=13cos2πt-π4

1013=cos2πt-π4

2πt-π4=cos-11013

t=12πcos-11013+π4=12πcos-11013+18

t=0.6932π+18=0.1103+0.125=0.235 Sec

t=235 ms

So, it can be seen that the instance of time at which the current is +10A is 235 milliseconds.

* The instantaneous value of a power signal may be described by:

12∠3π8 [W]

Find the magnitude of horizontal and vertical components of the signal.

Solution:

As this Power has been expressed in Polar form, with the magnitude of 12VA as the Apparent power and is subjected at an angle of 3π8 from the horizontal axis or the Real Power. We can calculate the horizontal (Real Power) and vertical (Reactive Power) components by using the method of converting Polar form into Rectangular form. The method is described as under:

r∠θ=rcosθ+jsinθ

4796155383286000Now let’s calculate the magnitudes of components:

Horizontal Component=12×cos3π8=4.59

Vertical Component=12×sin3π8=11.08

So, we can write the components in Cartesian or Rectangular form as under:

12∠3π8=4.59+j11.08

So, the Real Power component will be 4.59W and the Reactive Power component will 11.08VAR.

Task-2

* A resistor R is connected in series with an inductor L, an AC current i will flow this RL combination, causing a voltage VR of 30V to be developed across resistor and a voltage VL of 40V to be developed across inductor.

Assuming that VR is in phase with current i and VL leads VR by 90o, draw a vector diagram for this arrangement and then calculate the magnitude of the resultant voltage across the whole RL combination.

Solution:

31464854252990031712381615548Let the circuit be as shown in the figure below having resistor and inductor connected in series with the each other. The current i will flow through the circuit as both are in series s same current will flow through both R and L. The voltage drop can be different for both these components.

312955649345370Now, let’s draw the Vector diagram for the circuit. As we know the voltage and the current are in same phase or the phase angle between voltage and current is zero. So, we can express the vector diagram for this component as in figure showing current and voltage in phase.

In case of inductor, the voltage leads the current by 90o or the voltage attains its maximum value and zero value 90o before the current. This can be shown as in the figure.

Now for combined RL circuit, we can follow steps to draw the vector diagram:

3181937top0Step- I. In case of series RL circuit, resistor and inductor are connected in series, so current flowing through both the elements is same i.e. iR = iL = i. So, we can take current as reference and draw it on the horizontal axis as shown in diagram.

Step- II. In case of resistor, both voltage and current are in same phase. So, we can draw the voltage vector VR along same axis or direction to that of current vector.

rightcenter0Step- III. As in case of inductor, voltage leads the current by 90o, so we can draw VL perpendicular to current vector.

Step- IV. Now as we have two voltages VR and VL. So, we can draw the resultant vector VG of these two voltages as shown in the figure above. Such as,

VG2=VR2+VL2

So, we can calculate the combined voltage across the RL circuit as under:

VG=VR2+VL2=302+402=50V

We can calculate the phase angle of the right-angle triangle as under:

θ=tan-1VLVR=tan-14030=53o

So, it has been observed that when we combine both resistance and inductor, the phase angle of a series RL circuit will be between 0o to 90o.

* A current carrying filament is subjected to a strong magnetic field within an experimental chamber. It is required to find the force on the filament. The current may be mode

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