2
\$\begingroup\$

enter image description here

I'm reading a Floyd electronics book and working through some problems. This one has me stumped however. I already know the answer as it's given in the book (110k). However, I don't know how to actually solve the problem. I've tried approaching it from a few different angles, but none of them seem to work as I'm always short of a variable. Can anyone advise?

\$\endgroup\$
  • 1
    \$\begingroup\$ 1) Write out all the equations. 2) Substitute until you have a single unknown. 3) Solve for that unknown. 4) Go back to step 2 until you're out of unknowns. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 27 '17 at 0:33
  • \$\begingroup\$ I've tried that, but I always seem to have more than one unknown. I've tried the current divider formula, but I'm short of both Itotal and R2. I can't calculate the voltage across any of the resistors either as I don't know the total resistance. \$\endgroup\$ – esm Sep 27 '17 at 0:36
  • \$\begingroup\$ What have you tried? Can you write all the equations you've found in the circuits? \$\endgroup\$ – Oskar Skog Sep 27 '17 at 4:51
  • \$\begingroup\$ It's OK, I've got it now. The problem initially appeared to be a catch-22 as there was so much information 'missing'. I knew that if I had the total current, I could use KCL and get the current through R1 and thus the voltage. Likewise, if I knew the voltage across R3, I would know the voltage across R2 from KVL. The problem is that I didn't know how to find any of them. Once I've got the current through R1, it's trivial to solve. \$\endgroup\$ – esm Sep 27 '17 at 15:12
2
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

For the top resistors (all resistances calculated in 'k'): $$ V_1 = 47 x \tag 1 $$

For the bottom resistor: $$ V_3 = 33 (x + 1) \tag 2 $$

From voltage drop across R1 and R3:

$$ V_1 + V_3 = 220 \tag 3 $$

$$ 47x + 33 (x+1) = 220 \tag 4 $$

$$ 80x = 220 - 33 \tag 5 $$

$$ x = \frac {187}{80} \tag 6 $$

Now you can work out V1 and therefore R2.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your time on this. I didn't realise that it was just basic algebra. Perhaps I over-analysed the problem as usual. \$\endgroup\$ – esm Sep 27 '17 at 1:56
2
\$\begingroup\$

Here's the same schematic, drawn a little differently:

schematic

simulate this circuit – Schematic created using CircuitLab

You know that \$V_X=R_3\cdot I_3=R_3\cdot\left(I_1+1\:\textrm{mA}\right)\$.

Moving \$V_X\$ downward means less current in \$R_3\$ but also more current in \$R_1\$ that needs to go through \$R_3\$. Moving \$V_X\$ upward means more current in \$R_3\$ but also less current in \$R_1\$ to support that need for current in \$R_3\$. So this suggests that there is some middle value of \$V_X\$ that will be "just right."

A moment of thought and I think you also know that the current in \$R_1\$ is \$I_1=\frac{220\:\textrm{V}-V_X}{R_1}\$. That gives you the two bits of information you need.

Can you solve it now?

\$\endgroup\$
  • \$\begingroup\$ Yes, I understand now and I can derive the proper answer from the previous example given. It was the initial stage of finding I1 that I couldn't solve. R2 is easy to work out once I've got that. It was more a failure at basic maths than electronics. \$\endgroup\$ – esm Sep 27 '17 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.