Creo haber leído sobre esto..Un “interruptor” yo diría
Answer:
my_mul:
.globl my_mul
my_mul:
//Multiply X0 and X1
// Does not handle negative X1!
// Note : This is an in efficient way to multipy!
SUB SP, SP, 16 //make room for X19 on the stack
STUR X19, [SP, 0] //push X19
ADD X19, X1, XZR //set X19 equal to X1
ADD X9 , XZR , XZR //set X9 to 0
mult_loop:
CBZ X19, mult_eol
ADD X9, X9, X0
SUB X19, X19, 1
B mult_loop
mult_eol:
LDUR X19, [SP, 0]
ADD X0, X9, XZR // Move X9 to X0 to return
ADD SP, SP, 16 // reset the stack
BR X30
Explanation:
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
