circuit analysiscircuit-designpower electronics
I am trying to find the thevenin resistance value and the Vth for maximum power transfer. I have tried to find Rth by calculating R3 and R2 as parallel resistance and same for R4 and R5 and then adding them with R1 but I am not sure if I have done it the right way. For Vth I have tried using KVL but I think I have calculated wrong current values using current divider rule. I calculated the total current by diving V1 and R1. I hope someone can help. I have attached the picture of the circuit.
When you remove the load, you end up with:
simulate this circuit – Schematic created using CircuitLab
It follows that \$I_y=0\$ since it is an open circuit.
$$I_x=\dfrac{9\text{V}}{4\Omega+4\Omega+6\Omega}=0.643\text{A}$$
The potential difference between the terminals A and B is the \$V_{\text{TH}}\$. This is just the voltage across the farthest to the right \$4\Omega\$ resistor.
Then
$$V_{oc}=V_{\text{TH}}=(0.643\text{A})(4\Omega)=2.57\text{V} $$
You need to find \$R_{\text{TH}}\$. Short circuit the 9V voltage source and find the equivalent resistance at the A and B terminals. That would be:
$$ R_{\text{TH}}=((4\Omega+6\Omega)||4\Omega)+5\Omega=7.86\Omega$$
With that, you can find \$I_2\$.
There are several ways to reduce a net of sources and resistors to its Thevenin equivalent. One way is like what you started to do. Find the open circuit voltage and the short circuit current.
Another way is to keep simplifying the circuit in steps. Take a source and two resistors and find its Thevenin equivalent on its own. Then replace the three parts with the two Thevenin equivalent parts in the circuit. Keep doing this until the circuit is a single voltage source in series with a single resistor.
However, in all cases you should start by drawing the circuit in a comprehensible way. Problems like this are often deliberately drawn to obscure the circuit. Never let that get in your way. Redraw the schematic properly with power voltages descending down the page, logical flow left to right, etc. That should be your first step here too.
You should always draw circuits properly before asking others to look at them. This web site is no exception.
Best Answer
I post this as an answer so I can plot a circuit equivalent.
simulate this circuit – Schematic created using CircuitLab
So. There we can see the same circuit (without the supply, we want to calculate R equivalent as viewed from your two nodes.
If R1 weren't there, your equivalent would be (R2+R4)//(R3+R5).
What happens if we add R1? Well, if the bridge is balanced, meaning both dividers provide the same voltage, R1 would be doing nothing. That's not the case here.
Let's make it the lazy way first. Wanna know the resistance? We can put a voltage source and check the current, or put a current source and check the voltage. Let's click simulate circuit and see.
Put 1A, we get 4.713V. That would mean 4.713ohms.
Put 1V we get 212.2mA, as expected 4.713 ohms.