Answer:
- The resistance of the circuit is 1250 ohms
- The inductance of the circuit is 0.063 mH.
Explanation:
Given;
current at resonance, I = 0.2 mA
applied voltage, V = 250 mV
resonance frequency, f₀ = 100 kHz
capacitance of the circuit, C = 0.04 μF
At resonance, capacitive reactance (
) is equal to inductive reactance (
),
Where;
R is the resistance of the circuit, calculated as;

The inductive reactance is calculated as;

The inductance is calculated as;

Complete Question:
Find the capacitive reactance of a 0.1 microfarad capacitor with frequency 50Hz and at 200Hz.
Answer:
I. Capacitive reactance = 31826.86 Ohms.
II. Capacitive reactance = 7956.72 Ohms.
Explanation:
<u>Given the following data;</u>
Capacitance = 0.1 uF = 0.0000001 Farad
Frequency = 50 Hz and 200 Hz
To find the capacitive reactance;
Mathematically, the capacitive reactance of an electronic circuit is given by the formula;

Where;
- Xc is capacitive reactance.
- f is the frequency.
- c is the capacitance.
Substituting into the formula, we have;
I. At frequency, f = 50 Hz




II. At frequency, f = 200 Hz




Answer:
Hello your question is incomplete attached below is the missing part and answer
options :
Effect A
Effect B
Effect C
Effect D
Effect AB
Effect AC
Effect AD
Effect BC
Effect BD
Effect CD
Answer :
A = significant
B = significant
C = Non-significant
D = Non-significant
AB = Non-significant
AC = significant
AD = Non-significant
BC = Non-significant
BD = Non-significant
CD = Non-significant
Explanation:
The dependent variable here is Time
Effect of A = significant
Effect of B = significant
Effect of C = Non-significant
Effect of D = Non-significant
Effect of AB = Non-significant
Effect of AC = significant
Effect of AD = Non-significant
Effect of BC = Non-significant
Effect of BD = Non-significant
Effect of CD = Non-significant
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Answer:
Outdoors
Explanation:
Construction workers perform outdoors.