Am I correct in re-drawing the unbalanced bridge as 2 voltage dividers in parallel connected by the 2.7K resistor? Removing RL leaves me a circuit with 3 loops. Calculating branch currents and applying them to find Vth does indeed give me 7.6922 Volts.
But, how do I calculate Rth for this circuit? If I short out the 9V supply, it (I think) leaves me with the 1.5K and 3.3K in parallel. Is this correct? The 2.7K and 1K being in series produces a 3.7K that will be in parallel with the equivalent of the 1.5K and 3.3K? Now matter what I do I cannot come up with the textbook answer of Rth=766.84 ohms. I come up with roughly 806 ohms.
I'm obviously reading the resistors series/parallel combinations incorrectly. Can someone help me to correctly deduce Rth?
But, how do I calculate Rth for this circuit? If I short out the 9V supply, it (I think) leaves me with the 1.5K and 3.3K in parallel. Is this correct? The 2.7K and 1K being in series produces a 3.7K that will be in parallel with the equivalent of the 1.5K and 3.3K? Now matter what I do I cannot come up with the textbook answer of Rth=766.84 ohms. I come up with roughly 806 ohms.
I'm obviously reading the resistors series/parallel combinations incorrectly. Can someone help me to correctly deduce Rth?