I'm new to electronic circuit, I'm solving some problems in the book and come across this one, but I cannot calculate the value of Rs and Rp according to the info provided. Can someone help me with this, I solved equations using Thevenin but there is no solution for the value of those two resistors (sorry for my bad English, I'm Vietnamese ) enter image description here


There are two unknowns: \$R_s\$ and \$R_p\$. I.e. you need two equations.

  • 1st equation:
    \$V_g\$ with \$R_g\$ and \$R_s\$ form a Thevenin source with \$V_{Th1} = V_g\$ and \$R_{Th1} = R_g + R_s\$

    Transform this Thevenin source into a Norton source: \$R_{N1} = R_{Th1} = R_g + R_s\$.

    Include \$R_p\$ to form a new Norton source whose \$R_{N2} = R_{N1} || R_p\$.

    Transform it back to a Thevenin source \$R_{Th2} = R_{N2} = R_{N1} || R_p\$.

    The resistance of the final Thevenins source is supposed to equal that of the starting Thevenin source, i.e.

    \$R_{Th1} = R_{Th2}\$ i.e.
    \$R_g = R_{N1} || R_p = (R_g + R_s) || R_p\$

  • 2nd equation:
    You get from \$V_g / V_o = 0.125\$
    It depends how \$V_o\$ is defined (I think the problem is not clear about this): either with \$R_L\$ attached or without \$R_L\$ connected. I'm not sure which one is the case (the diagramm actually says with \$R_L\$; but \$R_L\$'s value is not given).
    In either case use the voltage divider formula.

  • \$\begingroup\$ but the problem says Req = 100, and this Req is the resistor of Attenuator. Does it mean that when this Attenuator is not attached to the circuit yet, then Req = Rs + Rp = 100 ? \$\endgroup\$ Oct 18 '17 at 13:40
  • \$\begingroup\$ @Curd I think the situation is absolutely clear. In the schematic diagram, the load \$R_L\$ is clearly attached to the output of the attenuator. So, both the \$R_s\$ and \$R_p\$ will be functions of \$R_L\$. \$\endgroup\$
    – Eric Best
    Oct 18 '17 at 13:55
  • \$\begingroup\$ @Uvuvwevwevwe Onyetenyevwe Ugwe No, \$R_{eq}\$ equals \$R_p||(R_s+R_g)\$ and the second condition says \$R_g = R_{eq}\$. \$\endgroup\$
    – Eric Best
    Oct 18 '17 at 14:03
  • \$\begingroup\$ That's exactly right, but the diagram definitely makes it look like some (possibly series) combination of just Rp and Rs which is equal to 100. \$\endgroup\$ Oct 18 '17 at 15:28
  • 1
    \$\begingroup\$ Thank god for user name autocompletion \$\endgroup\$
    – pipe
    Oct 18 '17 at 17:23

We can write:

\$ \frac{V_0}{V_g}=\frac{R_p||R_L}{R_p||R_L+R_s+R_g} \>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>(1) \$


\$ R_g=R_{eq}=R_p||(R_s+R_g) \> \> \> \> \>\>(2)\$

Let \$ \frac{V_0}{V_g}=0.125=k \$,

then, from the above equations, we can derive that

\$ R_p = (1+\frac {R_g}{R_s})R_g=(1+\frac {100}{R_s})100\$


\$ R_s = \frac{(1-k)R_L*R_p}{k(R_p+R_L)}-100= \frac{(1-k)R_L*(1+\frac {100}{R_s})100}{k((1+\frac {100}{R_s})100+R_L)}-100 \$

Solve the last two equations for \$R_s\$ and \$R_p\$, substitute \$k\$, and...

that's it!

Note: Both \$ R_s\$ and \$R_p\$ are functions of the load resistance \$R_L\$, of course.

Well, later I solved it and the general solution (including an optional value of \$ R_g\$ and \$\frac{V_0}{V_g}\$) is as follows:

\$ R_s=\frac{R_L \cdot R_g}{k(R_L+R_g)}-R_g=\frac{R_L||R_g}{k}-R_g =\frac{V_g}{V_0} \cdot (R_L||R_g)-R_g\$

\$ R_p=\frac{1}{\frac{1}{R_g}-\frac{k}{\frac{1}{R_g}+\frac{1}{R_L}}}=\frac{1}{\frac{1}{R_g}-\frac{k}{R_L||R_g}}=\frac{1}{\frac{1}{R_g}-\frac{V_0}{V_s} \cdot \frac{1}{R_L||R_g}}\$

All this holds if both results \$R_p\$ and \$R_s\$ are greater than 0 and the denominator in the \$R_p\$ formula is different from 0, of course. The condition for it is obvious from the formulae.

Continued on Oct 19, 2017

The above-mentioned condition is \$R_L>\frac{k}{1-k} \cdot R_g\$ and the specific results for our case (\$k=0.125\$ and \$R_g=100\$):

\$ R_s = \frac{875R_L-12500}{1.25R_L+125}\$,

\$R_p = \frac{1000R_L}{8.75R_L-125}\$

under the following condition:



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