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Mathematics
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Qatar University
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Applied Mathematics for Engineers
Question 1.
The flow control circuit of a rig given by a equation CR (dV/dt) + V = E where V is the
voltage, the potential difference between the plates of capacitor C charged by steady
voltage E through a resistor R .
a. Solve the equation for V when t= 0 and V=0
b. Given E = 25V, C = 20 X 10
-6
F, R = 200 X 10
3
ohms and t = 3.0 s.
Answer:



  



 

 


Integrating both side:
 



 

 
Where K is a constant
As, capacitor is charged by voltage E, so 
 
 
Using this and simplifying further:
 

 
Taking exponential of both sides:
 

 

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a. As, for , so,



so,
 

 
 

 
  

  


b. Putting 

 
:
  


 

Question 2:
The refinery in Oman wants to test the concentration of impurities of oil, the impurities is
Controlled by an equation a (dc / dt) + Cm = b + dm, where a, b, d and m are
constants. Given C= c
0
when t = 0,
Solve the equation and show that C - c
0
e
(mt / a)
= {(b/m) + d} {1 e
(mt / a)
}
Answer:


    



  

    

Integrating both side:
    





 

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Applied Mathematics for Engineers Question 1. The flow control circuit of a rig given by a equation CR (dV/dt) + V = E where V is the voltage, the potential difference between the plates of capacitor C charged by steady voltage E through a resistor R . a. Solve the equation for V when t= 0 and V=0 b. Given E = 25V, C = 20 X 10-6 F, R = 200 X 103 ohms and t = 3.0 s. Answer: 𝑑𝑉 )+𝑉 = 𝐸 𝑑𝑡 𝑑𝑉 ⇒ 𝐶𝑅 ( ) = 𝐸 − 𝑉 𝑑𝑡 𝑑𝑉 𝑑𝑡 ⇒ = 𝐸 − 𝑉 𝐶𝑅 𝐶𝑅 ( Integrating both side: ∫ 1 1 𝑑𝑉 = ∫ 𝑑𝑡 𝐸−𝑉 𝐶𝑅 ⇒ − ln |𝐸 − 𝑉| = 𝑡 +𝐾 𝐶𝑅 Where K is a constant As, capacitor is charged by voltage E, so 𝐸 ≥ 𝑉, 𝑠𝑜 |𝐸 − 𝑉| = 𝐸 − 𝑉 Using this and simplifying further: ln(𝐸 − 𝑉) = − ( 𝑡 + 𝐾) 𝐶𝑅 Taking exponential of both sides: 𝐸 − 𝑉 = exp (− 𝑡 − 𝐾) 𝐶𝑅 a. As, 𝑉 = 0 for 𝑡 = 0, so, 𝐸 = exp(−𝐾) ⇒ 𝐾 = −ln (𝐸) so, 𝐸 − 𝑉 = exp (− 𝑡 + 𝑙𝑛𝐸) 𝐶𝑅 𝑡 + 𝑙𝑛𝐸) 𝐶𝑅 𝑡 ⇒ 𝑉 = 𝐸 − 𝐸 exp (− ) 𝐶𝑅 𝒕 ⇒ 𝑽 = 𝑬 (𝟏 − 𝐞𝐱𝐩 (− )) 𝑪𝑹 ⇒ 𝑉 = 𝐸 − exp (− b. Putting 𝐸 = 25𝑉, 𝐶 = 20 𝑋 10−6 𝐹, 𝑅 = 200 ∗ 103 𝑜ℎ𝑚𝑠 𝑎𝑛𝑑 𝑡 = 3.0 𝑠: 𝑉 = 25 (1 − exp (− 3 )) = 𝟏𝟑. 𝟏𝟗 𝑽 20 𝑋 10−6 ∗ 200 ∗ 103 Question 2: The refinery in Oman wants ...
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