# How to calculate value of R2 in this circuit?

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?

• 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. – Ignacio Vazquez-Abrams Sep 27 '17 at 0:33
• 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. – esm Sep 27 '17 at 0:36
• What have you tried? Can you write all the equations you've found in the circuits? – Oskar Skog Sep 27 '17 at 4:51
• 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. – esm Sep 27 '17 at 15:12

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.

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

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

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?

• 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. – esm Sep 27 '17 at 2:23