I am a newbie to electronics so I need some help. I was reading about Thevenin equivalent circuits and I am pretty confused. I have the question like this:
I have a 2V battery which draws a maximum of 10A. What's the Thevenin equivalent circuit of that?
So, the first thing I think I should to do is to find R using Ohm's law. So, R = 0.2 ohms.
And now I would just draw it.
Is Thevenin equivalent like this?
If not, could you please give me steps how to solve this?
When you say that the battery "draws a maximum of 10A" do you mean that when a true short circuit is applied to the battery, so that the terminal voltage is zero, that the supplied current is 10A? If so then your calculation is correct. If that is the maximum rated current you should draw from the battery then you need to know the voltage at the battery terminals when 10A is being drawn.
If you want to determine the Thevenin equivalent voltage and resistance without overloading the battery, then apply some known resistance \$R_L\$ and measure the output voltage as \$V_L\$. Measure the voltage without a load as \$V_{OC}\$. The voltage divider equation tells us that
\$ V_L = V_{OC} \frac{R_L}{R_{TH}\times R_L}\$
Solve for \$R_{TH}\$, and you know that \$V_{TH} = V_{OC}\$.
Ya you are right.Just mention the polarity of the DC source.
Additionally , the resistance=Voltage/(current flowing when the two terminals are shorted) i.e when the current is max, which in this case,as u said is 10A i.e R=2/10=0.2 Ohm Hence the answer.
