# Modeling the Transient Response of Thermal Circuits
## 3 Elementary Thermal Circuits
### Transient Response Model
#### 3.2.3 Thermal Resistances in Parallel  
Let us consider the thermal circuit of Figure 2c. Let us also use the first Kirchhoff law at the nodes where the temperature is being recorded, and at the two nodes where the heat power is split. Let us also use the second Kirchhoff law in the thermal element loops and at the loop that connects \(R_{1}\) and \(R_{2}\). Hence,  
\[Q =q_{1}+q_{2}, \tag{37}\] \[q_{1} =q_{3}+q_{4},\] (38) \[q_{2} =q_{5}+q_{6},\] (39) \[R_{1}q_{3} =R_{2}q_{5},\] (40) \[\frac{dq_{3}}{dt} =\frac{1}{R_{1}C_{1}}q_{4},\] (41) \[\frac{dq_{5}}{dt} =\frac{1}{R_{2}C_{2}}q_{6}. \tag{42}\]Two of these differential equations are linear, ordinary, of first order and coupled. To solve this system, let us use two new variables, \(u_{3}=\frac{dq_{3}}{dt}\) and \(u_{5}=\frac{dq_{5}}{dt}\). By differentiating all the above equations, by using these two new variables and by doing the necessary algebra, one can decouple the differential equations and arrive at  
\[\begin{bmatrix}\frac{du_{3}}{dt}\\ \frac{du_{5}}{dt}\end{bmatrix} =\begin{bmatrix}-\frac{1}{R_{1}(C_{1}+C_{2})}&-\frac{1}{R_{1}(C_ {1}+C_{2})}\\ -\frac{1}{R_{2}(C_{1}+C_{2})}&-\frac{1}{R_{2}(C_{1}+C_{2})}\end{bmatrix} \begin{bmatrix}u_{3}\\ u_{5}\end{bmatrix} \tag{43}\] \[=M\begin{bmatrix}u_{3}\\ u_{5}\end{bmatrix}, \tag{44}\]  
where M is a square matrix. From the nature of this last system, we know that \(u_{3}\) and \(u_{5}\) are linear combinations of two exponential functions with time constants given by the eigenvalues of the matrix \(M\). In that respect, \(M\) has the trivial eigenvalue \(\lambda_{0}=0\) and the eigenvalue \(\lambda_{1}=-\frac{1}{C^{\dagger}}\Big{(}\frac{1}{R_{1}}+\frac{1}{R_{2}}\Big{)}\), where \(C^{\dagger}=C_{1}+C_{2}\). The respective eigenvectors are  
\[v_{\lambda_{0}}=\begin{bmatrix}1\\ -1\end{bmatrix},v_{\lambda_{1}}=\begin{bmatrix}1\\ \frac{R_{1}}{R_{2}}\end{bmatrix}. \tag{45}\]  
Therefore,  
\[\begin{bmatrix}u_{3}\\ u_{5}\end{bmatrix}=A\begin{bmatrix}1\\ 1\end{bmatrix}+B\epsilon^{\lambda_{1}t}\begin{bmatrix}1\\ \frac{R_{1}}{R_{2}}\end{bmatrix}, \tag{46}\]  
where \(A\) and \(B\) are constants. Hence, one can obtain:  
\[q_{4}(t) =R_{1}C_{1}Be^{\lambda_{1}t}+D, \tag{47}\] \[q_{6}(t) =R_{2}C_{2}Be^{\lambda_{1}t}+E, \tag{48}\]  
where \(D\) and \(E\) are also constants. Since \(q_{4}(t\rightarrow\infty)=0\), then \(D=0\). The same applies to \(q_{6}\), so that \(E=0\). Since \(q_{3}\) and \(q_{5}\) are initially \(0\), (open circuit), and by the definition of thermal capacitance, we have:  
\[\frac{q_{4}}{C_{1}}=\frac{Q-q_{4}}{C_{1}}. \tag{49}\]  
Therefore, \(q_{4}(t=0)=\frac{C_{1}}{C_{1}+C_{2}}Q\), which implies that \(B=\frac{Q}{C^{\dagger}R_{1}}\). Hence, one can finally arrive at  
\[q_{4} =\frac{C_{1}Q}{C^{\dagger}}e^{\frac{1}{C^{\dagger}}\Big{(}\frac{ 1}{R_{1}}+\frac{1}{R_{2}}\Big{)}t}, \tag{50}\] \[q_{6} =\frac{C_{2}Q}{C^{\dagger}}e^{\frac{1}{C^{\dagger}}\Big{(}\frac{ 1}{R_{1}}+\frac{1}{R_{2}}\Big{)}t}. \tag{51}\]  
By Equation (41),  
\[q_{3}=\frac{1}{R_{1}C_{1}}\frac{C_{1}Q}{C^{\dagger}}\bigg{(}\int e^{\frac{1}{ C^{\dagger}}\Big{(}\frac{1}{R_{1}}+\frac{1}{R_{2}}\Big{)}t}dt+D^{\prime}\bigg{)}. \tag{52}\]Since \(q_{3}(t=0)=0\), then \(D^{\prime}=\frac{QR_{2}}{R_{1}+R_{2}}\). Hence, one finally arrives at:  
\[q_{3}(t)=\frac{QR_{2}}{R_{1}+R_{2}}\bigg{[}1-e^{\frac{1}{C^{\frac{1}{C}}}\big{(} \frac{1}{R_{1}}+\frac{1}{R_{2}}\big{)}t}\bigg{]}. \tag{53}\]  
The same approach can be used to find \(q_{5}\):  
\[q_{5}(t)=\frac{QR_{1}}{R_{1}+R_{2}}\bigg{[}1-e^{\frac{1}{C^{\frac{1}{C}}}\big{(} \frac{1}{R_{1}}+\frac{1}{R_{2}}\big{)}t}\bigg{]}. \tag{54}\]  
Since \(\Delta T(t)=R_{1}q_{3}=R_{2}q_{5}\),  
\[\Delta T(t)=\frac{QR_{1}R_{2}}{R_{1}+R_{2}}\bigg{[}1-e^{\frac{1}{C^{\frac{1}{C }}}\big{(}\frac{1}{R_{1}}+\frac{1}{R_{2}}\big{)}t}\bigg{]}. \tag{55}\]