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RE=RA*[1+alpha*(TE-TA)] -> TE = ?

 05.10.2015

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 #1
avatar+26256 
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RE=RA*[1+alpha*(TE-TA)] -> TE = ?

 

\(\small{ \begin{array}{rcll} R_E &=& R_A\cdot [1+\alpha \cdot (T_E-T_A)] \qquad &| \qquad : R_A \\ \dfrac{R_E}{R_A} &=& 1+\alpha \cdot (T_E-T_A) \qquad &| \qquad -1\\ \dfrac{R_E}{R_A}-1 &=& \alpha \cdot (T_E-T_A) \qquad &| \qquad :\alpha\\ \left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right) &=& T_E-T_A \qquad &| \qquad +T_A \\ T_A+\left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right) &=& T_E \\ \mathbf{T_E} & \mathbf{=} & \mathbf{T_A+\left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right)} \\ \mathbf{T_E} & \mathbf{=} & \mathbf{T_A+ \dfrac{R_E-R_A}{R_A\cdot \alpha} } \\ \end{array} }\)

 

laugh

 05.10.2015
 #1
avatar+26256 
+9
Beste Antwort

RE=RA*[1+alpha*(TE-TA)] -> TE = ?

 

\(\small{ \begin{array}{rcll} R_E &=& R_A\cdot [1+\alpha \cdot (T_E-T_A)] \qquad &| \qquad : R_A \\ \dfrac{R_E}{R_A} &=& 1+\alpha \cdot (T_E-T_A) \qquad &| \qquad -1\\ \dfrac{R_E}{R_A}-1 &=& \alpha \cdot (T_E-T_A) \qquad &| \qquad :\alpha\\ \left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right) &=& T_E-T_A \qquad &| \qquad +T_A \\ T_A+\left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right) &=& T_E \\ \mathbf{T_E} & \mathbf{=} & \mathbf{T_A+\left(\dfrac{R_E}{R_A}-1 \right)\cdot \left(\dfrac{1}{\alpha} \right)} \\ \mathbf{T_E} & \mathbf{=} & \mathbf{T_A+ \dfrac{R_E-R_A}{R_A\cdot \alpha} } \\ \end{array} }\)

 

laugh

heureka 05.10.2015

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