# how to calculate resistors' value?

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 )

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.

• 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 ? Oct 18 '17 at 13:40
• @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 $R_L$. Oct 18 '17 at 13:55
• @Uvuvwevwevwe Onyetenyevwe Ugwe No, $R_{eq}$ equals $R_p||(R_s+R_g)$ and the second condition says $R_g = R_{eq}$. Oct 18 '17 at 14:03
• 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. Oct 18 '17 at 15:28
• Thank god for user name autocompletion
– pipe
Oct 18 '17 at 17:23

We can write:

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

and

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

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

then, from the above equations, we can derive that

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

and

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

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:

$R_L>\frac{R_g}{7}$