Answer:
Vx = 6.242 x 10raised to power 15
Vy = -6.242 x 10raised to power 15
Explanation:
from E = IVt
but V = IR from ohm's law and Q = It from faraday's first law
I = Q/t
E = Q/t x V x t = QV
hence, E =QV
V = E/Q
Answer:
- the capacity of the pump reduces by 35%.
- the head gets reduced by 57%.
the power consumption by the pump is reduced by 72%
Explanation:
the pump capacity is related to the speed as speed is reduces by 35%
so new speed is (100 - 35) = 65% of orginal speed
speed Q ∝ N ⇒ Q1/Q2 = N1/N2
Q2 = (N2/N1)Q1
Q2 = (65/100)Q1
which means that the capacity of the pump is also reduces by 35%.
the head in a pump is related by
H ∝ N² ⇒ H1/H2 = N1²/N2²
H2 = (N2N1)²H1
H2 = (65/100)²H1 = 0.4225H1
so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.
Now The power requirement of a pump is related as
P ∝ N³ ⇒ P1/P2 = N1³/N2³
P2 = (N2/N1)³P1
H2 = (65/100)²P1 = 0.274P1
So the reduction in power is 1 - 0.274 = 0.725 which is 72%
Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.
Answer:a
a) Vo/Vi = - 3.4
b) Vo/Vi = - 14.8
c) Vo/Vi = - 1000
Explanation:
a)
R1 = 17kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0
sin we know Va≈Vb=0
so
-Vi/5kΩ + -Vo/17kΩ = 0
Vo/Vi = - 17k/5k
Vo/Vi = -3.4
║Vo/Vi ║ = 3.4 ( negative sign phase inversion)
b)
R2 = 74kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
so
(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0
-Vi/5kΩ + -Vo/74kΩ = 0
Vo/Vi = - 74kΩ/5kΩ
Vo/Vi = - 14.8
║Vo/Vi ║ = 14.8 ( negative sign phase inversion)
c)
Also for ideal op-amp
Va≈Vb=0 so Va=0
Now for position 3 we apply nodal analysis we got at position 1
(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )
so
-Vi/5kΩ + -Vo/5000kΩ = 0
Vo/Vi = - 5000kΩ/5kΩ
Vo/Vi = - 1000
║Vo/Vi ║ = 1000 ( negative sign phase inversion)