Magnitudes of the currents are i1= 0.00104A , i2= 0.003641A , i3= 0.00508A.
Explanation:
Using superposition theorem,
remove the E1 voltage supply source and calculate resistance across it.
From the circuit given, as the resistances R1 and R2 are parallel
using the formula R1//R2=(R1.R2/(R1+R3)
V1= ((r1||r2)/(r1+r2||r3))*E1
v1 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*10v
v1= 2.3v
v2 = ((r1||r2)/(r1+r2||r3))*E2
v2 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*5v
v2= 1.161v
V = V1+V2
=> 2.3 + 1.161
=> 3.461v
Magnitudes of the currents can be found by i=v/r
i1 = v/r1
=> 3.461/3.3kΩ
=>0.00104A
i2=2.89/1kΩ
=>0.003461A
i3=2.89/680Ω
=> 0.00508A.
Therefore the magnitudes of the currents are i1, i2, and i3.
Answer:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Explanation:
A cubic has the form:
y = ax³ + bx² + cx + d
Given four points, we can write a system of equations:
1 = a + b + c + d
1/2 = 8a + 4b + 2c + d
1/3 = 27a + 9b + 3c + d
1/4 = 64a + 16b + 4c + d
Solving this algebraically would be time-consuming, but we can use matrices to make it easy.
First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:
Now we multiply by the solution matrix (again using a calculator):
So the cubic is:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Answer:
Pneumatics use easily-compressible gas like air or pure gas. Meanwhile, hydraulics utilize relatively-incompressible liquid media like mineral oil, ethylene glycol, water, synthetic types, or high temperature fire-resistant fluids to make power transmission possible
Explanation: