# Question 026bf

May 18, 2017

$13.85 \Omega$

#### 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 ${R}_{\text{th}}$ is Thevenin’s equivalent resistance which is equal to

The resistance across A and B gives the value of Thevenin’s resistance or ${R}_{\text{th}}$

May 18, 2017

${V}_{\text{Thevenin}} \approx 6.5 V$
${R}_{\text{Thevenin}} = 19.3 \Omega$

#### 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:

$\left({R}_{1} + {R}_{3}\right) {I}_{1} - {R}_{3} {I}_{2} = {V}_{1} \text{ [1]}$

Summing the voltages around the second window:

$- {R}_{3} {I}_{1} + \left({R}_{3} + {R}_{2} + {R}_{5}\right) {I}_{2} = 0 \text{ [2]}$

Substituting values:

$15 \Omega {I}_{1} - 5 \Omega {I}_{2} = 69 V \text{ [3]}$
$- 5 \Omega {I}_{1} + 23 \Omega {I}_{2} = 0 \text{ [4]}$

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

$64 \Omega {I}_{2} = 69 V$

${I}_{2} = \frac{69}{64} A$

${V}_{{R}_{5}} = {I}_{2} {R}_{5}$

${V}_{{R}_{5}} = \left(\frac{69}{64} A\right) \left(6 \Omega\right)$

${V}_{{R}_{5}} \approx 6.5 V$

Because no current flows through ${R}_{4}$ with ${R}_{L}$ removed, the voltage across ${R}_{5}$ is the Thevenin equivalent voltage:

${V}_{\text{Thevenin}} \approx 6.5 V$

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)))#

${R}_{\text{Thevenin}} = 19.3 \Omega$