Question

Thevenin Equivalents (circuits) Question

Answer 1

http://prntscr.com/cjmct5

Answer 2

@radar

Answer 3

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

So looking from R_load, the 1K and 2K are in parallel, and they're in parallel with the 3K
\[\large \frac{ 1 }{ R_{Th} } = \frac{ 1 }{ 1000 }+\frac{ 1 }{ 2000 }+\frac{ 1 }{ 3000 }\]And for V_th, just find the voltage across the 3K resistor as you would with any circuit (imagine the R_load is not even there).

Answer 4

To find the voltage across the 3K, first find the total R for the circuit. The 1K and 2K are in parallel, and in series with the 3K. So R_eq for the entire circuit
\[\large R_{eq }= 3000+\frac{ 1 }{ \frac{ 1 }{ 1000 }+ \frac{ 1 }{ 2000 } }\]Then use V=IR, where V = 5, so \[\large V = I R_{eq}\]to find I

Then use V = IR again, where R = 3000 and I is the value you just found. The value of V should be the V_th