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The question wants me to calculate that what is the \$R_L\$ value when we have the maximum \$P_{R_L}\$ from this circuit

schematic

simulate this circuit – Schematic created using CircuitLab

My thinking

When we are calculating the equivalent resistance,we can see the original circuit as the circuit in the left hand below ,and the left hand circuit is equal to the right hand circuit,by the way,\$R_{10}=R_1//R_3,\$ and \$R_{12}=R_2//R_4,\$

schematic

simulate this circuit

So according to \$P_{R_L}=I^2 R_L=(\frac{60}{2.1+R_L+0.9})^2R_L=\frac{60^2}{(3+R_L)^2}R_L\$,and use the Maximum Power Transfer Theorem,we can know \$R_L=3Ω\$, we can have the maximum \$P_{R_L}\$ from this circuit.

So now we can know the voltage of \$a\$ point,\$V_a\$, and the voltage of \$b\$ point,\$V_b\$ is as below

\$V_a=60\frac{3+0.9}{2.1+3+0.9}=39V\$

\$V_b=60\frac{0.9}{2.1+3+0.9}=9V\$

So the voltage of the loading resistor is \$39-9=30\$,so the \$P_{R_L}=\frac{V^2}{R_L}=\frac{30^2}{3}=300W \$

However,the answer shows me

Answer:

schematic

simulate this circuit

The \$V_{th}=60\frac{3}{7+3}-60\frac{1}{9+1}=12\$,and \$R_{th}=(7//3)+(9//1)=3 \$ ,so \$P_{R_L}=\frac{V^2}{4R_{th}}=12\$

I found that the main difference between solution and my thinking is that our \$V_{th}\$ are not the same ,my \$V_{th}\$ is the voltage of the loading resistor=30V,but solution's is \$12V\$ ,Why can't the voltage of loading resistor be calculated as i think??where am i wrong?

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You correctly found the Thevenin resistance of 3 ohm but then it went wrong when you assumed all the resistors were in series with 60 volts. The 3 ohms calculated is also the load you need to apply between A and B to get maximum power transfer too.

Why can't the voltage of loading resistor be calculated as i think??where am i wrong?

Consider that this can be solved by "splitting the voltage source" covered in E5 is this short document: -

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And that circuit becomes the one below when you convert the potential dividers on each supply and, calculate the new voltage source value for each side: -

enter image description here

So, the difference voltage between the two sources is 12 volts and it is this 12 volts that drives current through the 3 ohm resistor connected across A and B. So, that's: -

$$I = \dfrac{12\text{ volts}}{2.1 + 3 + 0.9} = 2 \text{ amps}$$

So the maximum power into the Thevenin resistance of 3 ohms is 12 watts.

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  • \$\begingroup\$ i don't understand why can we change the voltage source from \$60V\$ to \$18V\$ and \$6V\$ when we combine two resistors which they are parallel connected together,can you explain it of this part more ? \$\endgroup\$ – shineele Apr 28 '20 at 8:47
  • \$\begingroup\$ In my top diagram, what is the unloaded voltage at point A? \$\endgroup\$ – Andy aka Apr 28 '20 at 8:54
  • \$\begingroup\$ the unloaded voltage at point A is 18V,but the resistor below the point A is 3Ω \$\endgroup\$ – shineele Apr 28 '20 at 8:58
  • \$\begingroup\$ @shineele When you mentally remove the unknown resistor for a moment, you are left with two resistor dividers. The two separate Thevenin voltages at each of their shared nodes will be a divided voltage value, too. Not the full 60 V. Don't you see that? \$\endgroup\$ – jonk Apr 28 '20 at 8:58
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    \$\begingroup\$ @shineele correct 18 volts - and that potential divider that produces 18 volts is equivalent to an 18 volt source in series with a 2.1 ohm resistor. That is the point about converting a voltage source with two resistors to a (new but equivalent) voltage source with a single resistor. \$\endgroup\$ – Andy aka Apr 28 '20 at 9:11

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