Question

Question #026bf

Answer 1

`13.85Omega`

Explanation:

To find out the resultant Resistance please refer the Image below which gives better illustration about the diagram

now instead of directly substituting the values we we use there names .
and `R_("th")` is Thevenin’s equivalent resistance which is equal to

The resistance across A and B gives the value of Thevenin’s resistance or `R_("th")`

Answer 2

`V_"Thevenin"~~ 6.5V`
`R_"Thevenin" = 19.3Omega`

Explanation:

To find the Thevenin Voltage, you remove the load and then compute the open circuit voltage across the two points

Summing the voltages around the first window:

`(R_1+ R_3)I_1-R_3I_2 = V_1" [1]"`

Summing the voltages around the second window:

`-R_3I_1+ (R_3+R_2+R_5)I_2 = 0" [2]"`

Substituting values:

`15OmegaI_1-5OmegaI_2 = 69V" [3]"`
`-5OmegaI_1+ 23OmegaI_2 = 0" [4]"`

Multiply equation [4] by 3 and add to equation [3]:

`64OmegaI_2=69V`

`I_2=69/64A`

`V_(R_5)= I_2R_5`

`V_(R_5)= (69/64A)(6Omega)`

`V_(R_5)~~ 6.5V`

Because no current flows through `R_4` with `R_L` removed, the voltage across `R_5` is the Thevenin equivalent voltage:

`V_"Thevenin"~~ 6.5V`

To find the Thevenin equivalent resistance, we replace the voltage source with a wire and compute the equivalent resistance:

`R_"Thevenin" = R_4+ 1/(1/R_5+1/(R_2+1/(1/R_3+1/R_1))`

`R_"Thevenin" = 15Omega+ 1/(1/(6Omega)+1/(12Omega+1/(1/(5Omega)+1/(10Omega)))`

`R_"Thevenin" = 19.3Omega`