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Zorluk: ZorDirect, Inverse, Joint and Partial Variation

The total energy loss EE in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency ff in Hz of the alternating current. Given that E=120 JE = 120\text{ J} when f=10 Hzf = 10\text{ Hz}, and E=360 JE = 360\text{ J} when f=20 Hzf = 20\text{ Hz}, what is the value of EE in Joules when f=15 Hzf = 15\text{ Hz}?

Cevap: 220 J

Cevap

220
The relation describes a partial variation model E=c+kf2E = c + k f^2. Substituting the two given conditions (f=10,E=120f=10, E=120 and f=20,E=360f=20, E=360) yields the system of equations c+100k=120c + 100k = 120 and c+400k=360c + 400k = 360. Solving this system gives k=0.8k = 0.8 and c=40c = 40. Evaluating E=40+0.8(15)2E = 40 + 0.8(15)^2 results in 220 J220\text{ J}.

Adım Adım Çözüm

1
Write the general equation for partial variation involving a constant term and a term proportional to f2f^2
E=c+kf2E = c + k f^2
Partial variation consists of a sum of a constant component and a variable component.
2
Substitute the known conditions into the variation equation to create a system of linear equations
Equation 1: c+100k=120c + 100k = 120; Equation 2: c+400k=360c + 400k = 360
Two pairs of values are provided to determine the two unknown constants cc and kk.
3
Solve the system of simultaneous equations for kk and cc
k=0.8k = 0.8 and c=40c = 40
Subtracting Equation 1 from Equation 2 eliminates cc, allowing direct calculation of kk, after which cc is found by substitution.
4
Calculate the required value of EE when f=15f = 15
E=40+0.8(152)=40+0.8(225)=220E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220
Applying the discovered constants to the target frequency yields the final energy loss value.

Anahtar Kavram

Partial Variation and Simultaneous Equations
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