Penjumlahan dan pengurangan, bentuk polar harus diubah
ke bentuk rectangular.
1. 10 + j15 + (22 ∟ 30°) =
10 + j15 + [22×(cos 30) + j22×(sin 30)] =
10 + j15 + 22×(cos 30) + j22×(sin 30) =
10 + 22×(cos 30) + j15 + j22×(sin 30) = 29,05256 + j26
10 + j15 + (22 ∟ 30°) = 29,05256 + j26
10 + j15 + [22×(cos 30) + j22×(sin 30)] =
10 + j15 + 22×(cos 30) + j22×(sin 30) =
10 + 22×(cos 30) + j15 + j22×(sin 30) = 29,05256 + j26
10 + j15 + (22 ∟ 30°) = 29,05256 + j26
2. 21,3 − j32,5 − (9 ∟ 71,3°) =
21,3 − j32,5 − [9×(cos 71,3) + j9×(sin 71,3)] =
21,3 − j32,5 − 9×(cos 71,3) − j9×(sin 71,3) =
21,3 − 9×(cos 71,3) − j32,5 − j9×(sin 71,3) = 18,41448 − j41,02489
21,3 − j32,5 − (9 ∟ 71,3°) = 18,41448 − j41,02489
21,3 − j32,5 − [9×(cos 71,3) + j9×(sin 71,3)] =
21,3 − j32,5 − 9×(cos 71,3) − j9×(sin 71,3) =
21,3 − 9×(cos 71,3) − j32,5 − j9×(sin 71,3) = 18,41448 − j41,02489
21,3 − j32,5 − (9 ∟ 71,3°) = 18,41448 − j41,02489
3. 0,75 − j0,1 + (0,3 ∟ 40°) =
0,75 − j0,1 + [0,3×(cos 40) + j0,3×(sin 40)] =
0,75 − j0,1 + 0,3×(cos 40) + j0,3×(sin 40) =
0,75 + 0,3×(cos 40) − j0,1 + j0,3×(sin 40) = 0,97981 + j0,09284
0,75 − j0,1 + (0,3 ∟ 40°) = 0,97981 + j0,09284
0,75 − j0,1 + [0,3×(cos 40) + j0,3×(sin 40)] =
0,75 − j0,1 + 0,3×(cos 40) + j0,3×(sin 40) =
0,75 + 0,3×(cos 40) − j0,1 + j0,3×(sin 40) = 0,97981 + j0,09284
0,75 − j0,1 + (0,3 ∟ 40°) = 0,97981 + j0,09284
4. −30 − j32 − (64 ∟ 31°) =
−30 − j32 − [64×(cos 31) + j64×(sin 31)] =
−30 − j32 − 64×(cos 31) − j64×(sin 31)] =
−30 − 64×(cos 31) − j32 − j64×(sin 31)] = −84,85871 − j64,96244
−30 − j32 − (64 ∟ 31°) = −84,85871 − j64,96244
−30 − j32 − [64×(cos 31) + j64×(sin 31)] =
−30 − j32 − 64×(cos 31) − j64×(sin 31)] =
−30 − 64×(cos 31) − j32 − j64×(sin 31)] = −84,85871 − j64,96244
−30 − j32 − (64 ∟ 31°) = −84,85871 − j64,96244
5. −j65 + (−80 ∟ 15°) =
−j65 + [−80×(cos 15) − j80×(sin 15)] =
−j65 −80×(cos 15) − j80×(sin 15) =
−80×(cos 15) −j65 − j80×(sin 15) = −77,27407 −j85,70552
−j65 + (−80 ∟ 15°) = −77,27407 −j85,70552
−j65 + [−80×(cos 15) − j80×(sin 15)] =
−j65 −80×(cos 15) − j80×(sin 15) =
−80×(cos 15) −j65 − j80×(sin 15) = −77,27407 −j85,70552
−j65 + (−80 ∟ 15°) = −77,27407 −j85,70552
Perkalian dan pembagian, bentuk rectangular harus
diubah ke bentuk polar.
1. −2 + j6 × (5 ∟ 10°) =
[√(2²+6²) ∟ tan⁻¹(6÷−2)°] × (5 ∟ 10°) =
√(2²+6²) × 5 ∟ tan⁻¹(6÷−2)° + 10° = 31,62278 ∟ 118,43495°
−2 + j6 × (5 ∟ 10°) = 31,62278 ∟ 118,43495°
[√(2²+6²) ∟ tan⁻¹(6÷−2)°] × (5 ∟ 10°) =
√(2²+6²) × 5 ∟ tan⁻¹(6÷−2)° + 10° = 31,62278 ∟ 118,43495°
−2 + j6 × (5 ∟ 10°) = 31,62278 ∟ 118,43495°
2. 9 − j10 ÷ (12 ∟ 27,5°) =
[√(9²+10²) ∟ tan⁻¹(−10÷9)°] ÷ (12 ∟ 27,5°) =
√(9²+10²) ÷ 12 ∟ tan⁻¹(−10÷9)° − 27,5°) = 1,12114 ∟ −75,51279°
9 − j10 ÷ (12 ∟ 27,5°) = 1,12114 ∟ −75,51279°
[√(9²+10²) ∟ tan⁻¹(−10÷9)°] ÷ (12 ∟ 27,5°) =
√(9²+10²) ÷ 12 ∟ tan⁻¹(−10÷9)° − 27,5°) = 1,12114 ∟ −75,51279°
9 − j10 ÷ (12 ∟ 27,5°) = 1,12114 ∟ −75,51279°
3. 32 − j14 ÷ (−21 ∟ 63°) =
[√(32²+14²) ∟ tan⁻¹(−14÷32)°] ÷ (−21 ∟ 63°) =
√(32²+14²) ÷ −21 ∟ tan⁻¹(−14÷32)° − 63°) = −1,66326 ∟ −86,62938°
32 − j14 ÷ (−21 ∟ 63°) = −1,66326 ∟ −86,62938°
[√(32²+14²) ∟ tan⁻¹(−14÷32)°] ÷ (−21 ∟ 63°) =
√(32²+14²) ÷ −21 ∟ tan⁻¹(−14÷32)° − 63°) = −1,66326 ∟ −86,62938°
32 − j14 ÷ (−21 ∟ 63°) = −1,66326 ∟ −86,62938°
4. 43,2 + j21 ÷ (55 ∟ −30°) =
[√(43,2²+21²) ∟ tan⁻¹(21÷43,2)°] ÷ (55 ∟ −30°) =
√(43,2²+21²) ÷ 55 ∟ tan⁻¹(21÷43,2)° − (−30)° = 0,87334 ∟ 55,92490°
43,2 + j21 ÷ (55 ∟ −30°) = 0,87334 ∟ 55,92490°
[√(43,2²+21²) ∟ tan⁻¹(21÷43,2)°] ÷ (55 ∟ −30°) =
√(43,2²+21²) ÷ 55 ∟ tan⁻¹(21÷43,2)° − (−30)° = 0,87334 ∟ 55,92490°
43,2 + j21 ÷ (55 ∟ −30°) = 0,87334 ∟ 55,92490°
5. −j80 × (−33 ∟ 125°) =
[√(0²+80²) ∟ tan⁻¹(−80÷0)°] × (−33 ∟ 125°) =
√(0²+80²) × −33 ∟ tan⁻¹(−80÷0)° + 125° = −2640 ∟ 35°
−j80 × (−33 ∟ 125°) = −2640 ∟ 35°
[√(0²+80²) ∟ tan⁻¹(−80÷0)°] × (−33 ∟ 125°) =
√(0²+80²) × −33 ∟ tan⁻¹(−80÷0)° + 125° = −2640 ∟ 35°
−j80 × (−33 ∟ 125°) = −2640 ∟ 35°
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