Given the complex numbers A 1 = 6 ∠ 30 and A 2 = 4 + j 5 , (a) convert A 1 to rectangular form: (b) convert A 2 to polar and exponential form: (c) calculate A 3 = ( A 1 + A 2 ) , giving your answer in polar form: (d) calculate A 4 = A 1 A 2 , giving your answer in rectangular form: (e) calculate A 5 = A 1 / ( A 2 * ) giving your answer in exponential form.
Given the complex numbers A 1 = 6 ∠ 30 and A 2 = 4 + j 5 , (a) convert A 1 to rectangular form: (b) convert A 2 to polar and exponential form: (c) calculate A 3 = ( A 1 + A 2 ) , giving your answer in polar form: (d) calculate A 4 = A 1 A 2 , giving your answer in rectangular form: (e) calculate A 5 = A 1 / ( A 2 * ) giving your answer in exponential form.
Solution Summary: The author explains the rectangular form of a complex number A_1 and the exponential and polar forms of complex numbers.
Given the complex numbers
A
1
=
6
∠
30
and
A
2
=
4
+
j
5
, (a) convert
A
1
to rectangular form: (b) convert
A
2
to polar and exponential form: (c) calculate
A
3
=
(
A
1
+
A
2
)
, giving your answer in polar form: (d) calculate
A
4
=
A
1
A
2
, giving your answer in rectangular form: (e) calculate
A
5
=
A
1
/
(
A
2
*
)
giving your answer in exponential form.
I need help checking if its correct
-E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a)
R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b)
R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c)
(R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d)
Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be:
E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a)
-R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b)
-R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c)
R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d)
These two equations will be implemented to the matrix formula I = inv(A) * b
R11 R12
(R1 + R3 + R4)
-R4
-R4
R4 + R2
10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ V
Apply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) W
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