(a) 78.55Ω
(b) 720Ω
(c) 2181.8Ω
<h2>Explanation:</h2>(a) The inductive reactance, , is the opposition given to the flow of current through an inductor and it is given by;
= 2 π f L --------------------(i)
Where;
f = frequency
L = inductance
From the question;
f = 50.0Hz
L = 250mH = 0.25H
Take π = 3.142 and substitute these values into equation (i) as follows;
= 2 π (50.0) (0.25)
= 25(3.142)
= 78.55Ω
Therefore, the inductive reactance is 78.55Ω
(b) The capacitance reactance, , is the opposition given to the flow of current through a capacitor and it is given by;
= (2 π f C) ⁻ ¹ --------------------(ii)
Where;
f = frequency
C = capacitance
From the question;
f = 50.0Hz
C = 4.40μF = 4.40 x 10⁻⁶ F
Take π = 3.142 and substitute these values into equation (ii) as follows;
= [2 π (50.0) (4.40 x 10⁻⁶)] ⁻ ¹
= [440 x (3.142) x 10⁻⁶)] ⁻ ¹
= [1382.48 x 10⁻⁶] ⁻ ¹
= [1.382 x 10⁻³] ⁻ ¹
= 0.72 x 10³ Ω
= 720Ω
Therefore, the capacitive reactance is 720Ω
(c) Impedance, Z, is the ratio of maximum voltage, to maximum current,
, flowing through a circuit. i.e
Z = -------------------(iii)
From the question;
= 240V
= 110mA = 0.11A
Substitute these values into equation (iii) as follows;
Z =
Z = 2181.8Ω
Therefore, the impedance in the circuit is 2181.8Ω
Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = = 240 / 0.11 = 2181.82 ohms
We need to define the variables,
So,
Therefore, the probability that the repair time is more than 4 horus can be calculate as,
The probability that the repair time is more than 4 hours is 0.136
b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,
The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63
Answer:
The value of v2 in each case is:
A) V2=3v for only Vs1
B) V2=2v for only Vs2
C) V2=5v for both Vs1 and Vs2
Explanation:
In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.
Also, what the problem asks is the value V2 in each case, where:
If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.
In the first case we can use an equivalent resistance between R2 and R3:
And
In the second case we can use an equivalent resistance between R2 and (R1+R4):
And
If we consider both batteries:
