I have the following problem:
Consider the circuit below:
The component values are: \$R_1 =5000 \Omega\$, \$R_2 =1000 \Omega\$, \$V_z = 5 \text{V} \$.
When \$i_1\$ jumps from \$12 \text{mA} \$ to \$0 \text{mA}\$, what is the time constant for discharging the inductor immediately after?
\$ \tau = L/R_1\$
\$ \tau = L/R_2\$
\$ \tau = L/(R_1+R_2)\$
\$ \tau = L/(R_1||R_2)\$
Here are my thoughts.
Since current through an inductor can't change momentarily, \$i_L=12 \text{mA}\$. The current will run through the zener-diode, when \$V_z=5 \text{V}\$.
The zener-diode allows a current of \$i = \frac{5 \text{V}}{5000 \Omega}=1 \text{mA}\$ to run through \$R_1\$, while there flows \$11 \text{mA}\$ through the zener-diode. These two currents meet at the middle node, and causes \$ 12 \text{mA}\$ to flow through \$R_2\$.
So as far as I see it, the current flows through bots \$R_1\$ and \$R_2\$ back into the inductor. \$R_1\$ and \$R_2\$ are in parallel, so my guess it that the time constant is expressed as: \$ \tau = L/(R_1||R_2)\$.
HOWEVER, it turns out the correct answer is \$ \tau = L/R_2\$ which I don't quite understand.
Can someone explain to me why this is the case?