Answers and Discussions for
Contents
Question 2
Setting Up the Problem
The following information about the circuit model of the transformer may be obtained from the figure:
- Winding resistances: $ r_1 = r_2 = 1 \, \Omega $
- Winding leakage inductances: $ L_{\ell 1} = L_{\ell 2} = 0 $
- Magnetizing inductances: $ L_{m1} = L_{m2} = 1 \, \textrm{H} $
- Winding turns: $ N_1 = N_2 = 10 $
The following information about the circuit excitation and state of the transformer is given in the text (note that it is easier to avoid referred quantities because of how the information is given):
- Initial primary winding voltage: $ v_1(0) = 10 \, \textrm{V} $
- Secondary winding current (initially open circuited): $ i_2(t) = 0, \; 0 \leq t < 1 \, \textrm{s} $
- Primary winding current (becomes open circuited): $ i_1(t) = \begin{cases} 10 \left(1 - e^{-t}\right) \, \textrm{A} \; &0 \leq t < 1 \, \textrm{s} \\ 0 \; &t \geq 1 \, \textrm{s} \end{cases} $
- For convenience, it is given that $ i_1 \left(t = 1^- \, \textrm{s}\right) = 6.32 \, \textrm{A} $
- Secondary winding voltage (becomes short circuited): $ v_2(t) = 0, \; t \geq 1 \, \textrm{s} $
The electrical energy provided to the machine is denoted $ W_E $.
$ \begin{equation} W_E = \sum_{k = 1}^{N} \int v_k i_k \, dt \end{equation} $
From KVL, the voltage equations needed for the energy calculation are found ($ p $ denotes the Heaviside operator).
$ \begin{align} v_1 &= r_1 i_1 + p\lambda_1 \\ v_2 &= r_2 i_2 + p\lambda_2 \end{align} $
The flux linkage equations for a transformer are used. Recall that the magnetic flux linked to winding 1 due to a current in winding 2 is $ \Phi_{1,2} = L_{m1} \frac{N_2}{N_1} i_2 $.
$ \begin{align} \lambda_1 &= L_{\ell 1} i_1 + L_{m1} \left(i_1 + \frac{N_2}{N_1} i_2\right) \\ \lambda_2 &= L_{\ell 2} i_2 + L_{m2} \left(i_2 + \frac{N_1}{N_2} i_1\right) \end{align} $
Method 1: Exhaustive Calculation
The electrical energy supplied to this transformer for $ t \geq 1 \, \textrm{s} $ may be found by combining the voltage equations and flux linkage equations to the first equation for $ W_E $. The leakage terms are omitted since $ L_{\ell 1} = L_{\ell 2} = 0 $. $ \tau $ is used as a dummy variable for integrals.
$ \begin{align} W_E &= \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} r_1 \cancelto{0}{i_1^2} \, d\tau \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} L_{m1} \cancelto{0}{i_1} \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 0}^{1 \, \textrm{s}} \frac{N_2}{N_1} L_{m1} i_1 \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} \frac{N_2}{N_1} L_{m1} \cancelto{0}{i_1} \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} r_2 \cancelto{0}{i_2^2} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} L_{m2} \cancelto{0}{i_2} \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} L_{m2} i_2 \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 0}^{1 \, \textrm{s}} \frac{N_1}{N_2} L_{m2} \cancelto{0}{i_2} \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} \frac{N_1}{N_2} L_{m2} i_2 \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} \\ \end{align} $
All the zero terms can be removed, and time integrals with inductance are replaced with current integrals using the currents at the time bounds.
$ \begin{align} W_E &= \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{i_1 = 10 \, \textrm{A}}^{6.32 \, \textrm{A}} L_{m1} i_1 \, di_1 + \int_{i_2 = 0}^{0} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \, di_2 \\ &{} + \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau + \int_{i_2 = 0}^{i_2(t), \, t \geq 1 \, \textrm{s}} L_{m2} i_2 \, di_2 + \int_{i_1 = 0}^{0} \cancelto{1}{\frac{N_1}{N_2}} L_{m2} i_2 \, di_1 \end{align} $
Any integration between identical bounds will have a zero result. Because the secondary winding is short circuited $ v_2(t) = 0, \; t \geq 1 \, \textrm{s} $, it is further known that $ \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau + \int_{i_2 = 0}^{i_2(t), \, t \geq 1 \, \textrm{s}} L_{m2} i_2 \, di_2 = 0 $ as a consequence of the voltage equations. The electrical energy waveform must account for the period $ 0 \leq t < 1 \, \textrm{s} $ and revert to time integrals using $ \frac{di_1}{dt} = +10 e^{-t} \, \frac{\textrm{A}}{\textrm{s}} $.
$ \begin{equation} W_E(t) = \begin{cases} \int_{\tau = 0}^{t} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{t} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases} \end{equation} $
$ \begin{equation} \boxed{W_E(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $
Method 2: Fast Calculation
The electrical energy supplied to this transformer for $ t \geq 1 \, \textrm{s} $ may be found for each time period and each winding. $ \tau $ is used as a dummy variable for integrals.
$ \begin{equation} W_E = \int_{\tau = 0}^{1 \, \textrm{s}} v_1 i_1 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} v_2 \cancelto{0}{i_2} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} v_1 \cancelto{0}{i_1} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} \cancelto{0}{v_2} i_2 \, d\tau \end{equation} $
By removing the zero terms, it can be shown that nonzero electrical energy is only supplied to the transformer in winding 1 during the first second of time. An answer is readily obtained using the equation for $ v_1 $ (including $ \lambda_1 $), substituting the given equation for $ i_1(t) $, and removing the leakage terms since $ L_{\ell 1} = L_{\ell 2} = 0 $.
$ \begin{equation} W_E(t) = \begin{cases} \int_{\tau = 0}^{t} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{t} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau + \int_{\tau = 0}^{t} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \cancelto{0}{\frac{di_2}{d\tau}} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \cancelto{0}{\frac{di_2}{d\tau}} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases} \end{equation} $
Since winding 2 is open circuited during the first second of time, the time derivative of the winding 2 current is also zero. The final answer is obtained by removing terms, finding $ \frac{di_1}{dt} = +10 e^{-t} \, \frac{\textrm{A}}{\textrm{s}} $, and substituting parameters.
$ \begin{equation} \boxed{W_E(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $
