Is my circuit solution correct?

Question

^{simulate this circuit – Schematic created using CircuitLab}

I have to calculate the energy of the inductors magnetic field at t = 6 ms.

I am given: E = 50 V, R1 = 10 Ω, R2 = 15 Ω, R3 = 20 Ω and L = 0.1 H

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So at t < 0, the switch is closed and the current is flowing. In the parallel section, we get I1 and I2, since the inductor formula is time-dependent and it is t < 0 the inductor will have no voltage and will behave like a wire.

I calculate the total resistance in the parallel and add R3 for Rtotal;

$$ R_{total} = (\frac{1}{R_1} + \frac{1}{R_2})^{-1} + R_3 = 26 Ω $$

$$ I = \frac{U}{R_{total}} = \frac{50}{26} = 1.92 A $$

Now I can calculate the voltages in the circuit parts;

$$ U_{R3} = I * R_3 = 1.92 * 20 = 38.40 V $$

The voltage in parallel should be the same so UR1 = UR2 and together with UR3 should amount to 50 V, 50 - 38.40 = 11.60V in parallel.

At t = 0 the switch is open. The current stops flowing from the source and the depletion of the inductor starts.

It is a falling function the only voltage provider is the inductor;

$$ I = \frac{U_L}{R_1 + R_2} \rightarrow 1.92 = \frac{U_L}{25} \rightarrow U_L = 48 V $$

So since it is t = 0 things will remain the same;

$$ I_L = 1.92 \exp{\frac{-0}{\tau}} = 1.92 A $$

$$ U_L = 48 \exp{\frac{-0}{\tau}} = 48 V $$

At t = 6ms = 0.006 s;

$$ \tau = \frac{L}{R_{total}} = \frac{0.1}{25} = 0.004 $$

$$ I_L = 1.92\exp{\frac{-0.006}{0.004}} = 0.42 A $$

$$ U_L = 48\exp{\frac{-0.006}{0.004}} = 10.71 V $$

The energy of the inductor at t = 6 ms;

$$ W_L = \frac{Li^2}{2} = \frac{0.1 * 0.42^2}{2} = 0.008 J $$

Answer 1

At t < 0:

The switch is closed and the current is flowing. We assume that the system has reach the stationary phase. So, the inductor (its formula is time-dependent) the inductor will have no voltage and will behave like a wire.

^{simulate this circuit – Schematic created using CircuitLab}

You calculated the total resistance for this circuit:

$$ R_{total} = (\frac{1}{R_1} + \frac{1}{R_2})^{-1} + R_3 ≃ 26 Ω $$ Which leads to: $$ I_T = \frac{U}{R_{total}} = \frac{50}{26} ≃ 1.92 A $$ Now, you can calculate the voltages and currents through the circuit: $$ U_{R3} = I_T*R_{3} = 1.92 *20 ≃ 38.40V $$ The voltage in parallel should be the same so UR1 = UR2. Together with UR3 should amount to 50 V, UR1 = UR2 = 50 - 38.40 = 11.60V in parallel.

With ohm law, you can determine the currents going through the resistor and deduce the current going through the inductor:

$$ U_{R1} = R_{1} * i_{1} => I_{1} ≃ 1.15A $$ $$ U_{R2} = R_{2} * i_{2} => I_{2} ≃ 0.77A $$

Resistor R2 and inductor L are components in serie:

But $$ I_{R_2} = I_L ≃ 0.77A $$

At t = 0 :

The switch is open. The current stops flowing from the source and the depletion of the inductor starts.

^{simulate this circuit}

You will notice that I change the convention of the voltage for the resistor R1 in order to keep the voltage arrow of R1 in the "oposite direction" of the current arrow going through R1.

It is a falling function the only voltage provider is the inductor. It comes from the differential equation of the circuit:

Use the Kirchhoff law for meshes: $$ U_{L} + U_{R1} + U_{R2} = 0 $$ But, you know that, for an inductor, $$ U_{L} = L\frac{di(t)}{dt} $$ with i(t) the current going through the inductor !

The ohm law gives you: $$ U_{R1} = R_{1} * i(t) $$ and $$ U_{R2} = R_{2} * i(t) $$ This leads to: $$ L\frac{di(t)}{dt} + R_{1} * i(t) + R_{2} * i(t) = 0 $$ The solution of this differential equation is: $$ i(t) = cst*e^{ \frac{-t}{tau} } $$ with cst to be determined and tau the characteristic time of your circuit for t>0 with $$ tau = \frac{L}{R_{1}+R_{2}} = 0.004s $$ (this is math)

Now, (this is your mistake), you have to find cst, for that you know that the current flowing an inductor is continuous, so:

$$ i(t = 0^{-}) = i(t = 0^{+}) $$ Or $$ i(t = 0^{-}) = I_L ≃ 0.77A ≠ 1.92 A !$$

Consequently, you have: $$ i(t) = 0.77*e^{ \frac{-t}{0.004} } $$

Which leads you to t=6ms: $$ i(t) = 0.77*e^{ \frac{-0.006}{0.004} } ≃ 0.17A $$

So: The energy of the inductor at t = 6 ms is: $$ W(t=6ms) = \frac{L*(i(t=6ms))^2}{2} = \frac{0.1*0.17^2}{2} ≃ 0.0015J $$