Tutorial 1: Worked solutions
Electrical Engineering I · Module 92315 · Tutorial 1
Use these first tasks to get comfortable with SI prefixes, vector arithmetic and coordinate systems. Try each part before opening its worked steps.
Task 1: Convert units
Rewrite each quantity using a suitable SI prefix while keeping its physical value unchanged.
Connected topic: Electric charge and field quantities (§1.1) use SI units; prefixes are powers of ten attached to those units.
1a–f: Choose the prefix
Try first: move the power of ten into a prefix and check that the leading number is in a convenient range.
Attempted it? Open the worked steps
1a–f · Prefix conversions
1a. The decimal 0.000011 is 11 × 10−6. The prefix micro (μ) means 10−6:
C = 0.000011 F = 11 × 10−6 F
C = 11 μF
Check: 11 μF is 0.000011 F.
1b. Multiply the mantissas first: 47.47 × 105 = 4.747 × 106. Mega (M) means 106:
R = 4.747 MΩ
Check: 4.747 MΩ = 4,747,000 Ω = 47.47 × 100,000 Ω.
1c. Combine the decimal factor and the given power: 0.062 × 10−3 = 62 × 10−6. Micro means 10−6:
L = 62 μH
Check: 62 μH = 0.000062 H.
1d. Kilo (k) means 103, and 30,000 = 30 × 103:
U = 30 kV
Check: 30 kV = 30,000 V.
1e. 0.88 × 10−4 = 8.8 × 10−5 = 88 × 10−6. Micro means 10−6:
I = 88 μA
Check: 88 μA = 0.000088 A.
1f. 1265 × 106 = 1.265 × 109. Giga (G) means 109:
P = 1.265 GW
Check: 1.265 GW = 1,265,000,000 W.
Task 2: Cartesian coordinate system
The points are P1 = (2, −1, 2) and P2 = (1, 2, 3). The coordinates have no stated unit, so the vector results are given in coordinate units.
Connected topic: Electric field as a vector (§1.3) uses components, magnitudes and directions in the same Cartesian basis.
2a: Displacement from P1 to P2
Try first: subtract the starting point coordinates from the ending point coordinates, component by component.
Attempted it? Open the worked steps
2a · Displacement vector
A vector from point 1 to point 2 is the endpoint position minus the starting position:
r12 = rP2 − rP1
Subtract each matching coordinate:
r12 = (1 − 2, 2 − (−1), 3 − 2)
r12 = (−1, 3, 1)
r12 = −ex + 3ey + ez
Result: the displacement is (−1, 3, 1) coordinate units. The negative x-component means the motion is toward decreasing x; the y and z components are positive.
Check: adding the displacement to P1 gives (2, −1, 2) + (−1, 3, 1) = (1, 2, 3), the stated endpoint.
2b: Magnitude of r = −ex + 3ey + ez
Try first: square each perpendicular component, add them, then take the positive square root.
Attempted it? Open the worked steps
2b · Vector magnitude
For Cartesian components, the length follows from the three-dimensional Pythagorean theorem:
|r| = √rx2 + ry2 + rz2
Here the components are (−1, 3, 1), so:
|r| = √(−1)2 + 32 + 12
|r| = √11 ≈ 3.32
Result: |r| = √11 coordinate units.
Check: a magnitude cannot be negative, and its square is 1 + 9 + 1 = 11.
2c: Unit vector in the direction of r
Try first: divide the vector by its magnitude; this preserves its direction and makes its length one.
Attempted it? Open the worked steps
2c · Normalised direction
The unit vector is the original vector divided by its positive length:
er = r|r|
Divide the vector from 2a by its magnitude from 2b:
er = −ex + 3ey + ez√11
Result: er = (−1, 3, 1)/√11.
Check: its length is √(1 + 9 + 1)/11 = 1.
Task 3: Coordinate systems
Use the standard spherical convention in this course: φ is the azimuth in the xy-plane from +x, and ϑ is the polar angle down from +z.
Connected topic: Flux and Gauss’s law (§1.10) uses spherical coordinates for radial fields and spherical surfaces.
3a: Convert the position vector to spherical coordinates
Try first: find the distance from the origin, keep the given azimuth, and use the ratio of cylindrical radius to height to find the polar angle.
Attempted it? Open the worked steps
3a · Cylindrical to spherical
The cylindrical components are ρ = 5, z = 2, and φ = 45°. The spherical radius is the distance from the origin:
r = √ρ2 + z2 = √25 + 4 = √29
ϑ is measured from +z, so its tangent is opposite over adjacent = ρ/z. The point is above the xy-plane, hence ϑ is in the first quadrant:
ϑ = arctan(ρz) = arctan(52) ≈ 68.2°
The azimuth remains the supplied angle, φ = 45°. State the spherical coordinates by name so the angle order is clear:
r = √29
φ = 45°
ϑ ≈ 68.2°
As a position vector, r = rer. Since ϑ = arctan(5/2), sinϑ = 5/√29 and cosϑ = 2/√29, so the vector has no eϑ component:
r = √29 er
Check: its cylindrical projection is r sinϑ = 5 and its z-component is r cosϑ = 2.
3b: Complete the spherical unit-vector table
Try first: remember that φ locates the projection in the xy-plane and ϑ measures down from +z. At the z-axis, azimuth is undefined.
Attempted it? Open the worked steps
3b · Unit vectors and spherical angles
For a unit direction, take its projection onto each Cartesian basis vector:
er · ex = sinϑ cosφ
er · ey = sinϑ sinφ
er · ez = cosϑ
Evaluate these three components for each listed direction. At either pole (±ez), φ can be chosen arbitrarily because the projection into the xy-plane has zero length.
The completed entries are:
Cartesian unit vector φ ϑ ez any (the sheet’s *) 0° −ez any (the sheet’s *) 180° ex 0° 90° −ex 180° 90° ey 90° 90° −ey 270° 90° (ex + ey)/√2 45° 90° (ex − ez)/√2 0° 135° (ey + ez)/√2 90° 45° For the last three rows, substitute the angles into the unit-vector formula. For example, φ = 0° and ϑ = 135° gives (sin135°, 0, cos135°) = (1/√2, 0, −1/√2).
Check: every listed Cartesian vector has unit length. Angles differing by 360° represent the same azimuth, so −90° is equivalent to 270°.
Personal worked-solution guide based on the supplied EE1 Tutorial Task sheet, written in my own words; no tutorial text or figures are reproduced. Not an official FAU publication; check course details against current material. Made by Thanate Amornjiraporn (Focus).