# Why isn't $V_L$ not equal to $i_LR_L$ according to Thevenin's theorem?

According to Thevenin's Theorem $$I_L= \frac{V_{th}-V_L}{R_g}$$ whereas $$I_L$$= current through the load. $$V_{th}$$= The potential difference across the load resistance when the load resistance is disconnected. $$V_L$$=The potential difference across the load resistance when the load resistance is connected. $$R_g$$= The equivalent resistance of the curcuit between the load points. Now why can't we write $$V_L = i_LR_L$$ by the ohm's law?

## 1 Answer

Now why can't we write $$V_L=i_LR_L$$ by the superposition?

All you need is Ohm's law to write that (superposition is typically used for circuits with more than one source so I'm not sure why you're attempting to apply it here). But this is consistent with Thevenin's theorem isn't it?

The total series resistance is

$$R_{eq} = R_g + R_L$$

thus

$$I_L = \frac{V_{th}}{R_{eq}} = \frac{V_{th}}{R_g + R_L}$$

Now apply Ohm's law

$$I_L = \frac{V_{th}}{R_g + \frac{V_L}{I_L}}$$

and it follows that

$$I_L = \frac{V_{th} - V_L}{R_g}$$

• Yeah superposition isn't the thing. Sorry. Im editing. – Nobody recognizeable Apr 24 at 2:39