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How to find the thevenin equivalent resistance and voltage?

schematic

simulate this circuit – Schematic created using CircuitLab

The solution for calculating the thevenin resistance is

\$(5//20)+2+1+2=9\$,i can understand this methodm,just let the current source become open circuit,and let the voltage source become short circuit

However,i can't understand the solution for calculating the thevenin voltage

\$V_{th}=5 \times 2 +25-25\times \frac{20}{20+5}=15V\$

Can anyone tell me why can the author just write the \$V_{th}\$ formula above?

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  • \$\begingroup\$ What you have written is incorrect. The solution for calculating the thevenin resistance is: - $$(5||20)+2+1+2$$ \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 12:23
  • \$\begingroup\$ @Andyaka oh yes! 5Ω and 20Ω are parallel connected \$\endgroup\$
    – shineele
    Apr 25 '20 at 12:31
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schematic

simulate this circuit – Schematic created using CircuitLab

While measuring the Thevenin voltage you have to assume that load isn't connected. Now, Convert the current source into a voltage source. As no load is connected, no current will flow through resistor '\$R_1\$' and hence will cause no voltage drop.

Current will circulate through '\$R_2\$' and '\$R_3\$' because of the 25 volts battery. According to the voltage decider rule, the voltage at node '\$A\$' will be \$\frac{5}{(5+20)}\times 25 =5\$volts.

Now, applying KVL at the output loop, \$V_{th} = V_1 + V_2 = 10 + 5 = 15\$ volts

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  • \$\begingroup\$ @shineele, how did you write the fraction??? \$\endgroup\$
    – Sadat Rafi
    Apr 26 '20 at 9:29
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    \$\begingroup\$ use $frac{a}{b}$,and use \$ to replace $ \$\endgroup\$
    – shineele
    Apr 26 '20 at 10:10
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Simplify and then simplify again: -

enter image description here

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  • \$\begingroup\$ how do you know the voltage of lower part of circuit is 5V? the voltage of upper part is \$5A \times 2=10\$ ? \$\endgroup\$
    – shineele
    Apr 25 '20 at 12:47
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    \$\begingroup\$ The upper part is self contained i.e. the current can only flow in R2 and so you can ignore that and concentrate of 25 volts being potentially divided down by the ratio 5/(20 + 5). This makes the voltage across the 5 ohm resistor 5 volts. \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 12:49
  • \$\begingroup\$ You can always use good old fashioned KVL and KCL! \$\endgroup\$
    – Leoman12
    Apr 25 '20 at 13:42
  • \$\begingroup\$ @Leoman12 - I have never used either since being at college (1980s) believe it or not. Norton's and Thevenins plenty of times but Kirchhoff not once (knowingly). \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 13:51

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