R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
Read more about resistance here:
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Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Answer:
1700kJ/h.K
944.4kJ/h.R
944.4kJ/h.°F
Explanation:
Conversions for different temperature units are below:
1K = 1°C + 273K
1R = T(K) * 1.8
= (1°C + 273) * 1.8
1°F = (1°C * 1.8) + 32
Q/delta T = 1700kJ/h.°C
T (K) = 1700kJ/h.°C
= 1700kJ/K
T (R) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8R
= 944.4kJ/h.R
T (°F) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8°F
= 944.4kJ/h.°F
Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C
= 2°C
(273+5) - (273+3)
= 2 K
