How to Determine Unknown Variables Using Voltage Division

I left out the units.

total resistance \$R\$

\$R_1\$ (between a and c) is in parallel to the rest of the resistors. The rest being \$R_2\$ parallel to \$R_3\$ (between a and b) in row to \$R_4\$ in parallel to \$R_5\$ (between b and c)

The general rule for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel:

$$R_a||R_b=\frac{R_aR_b}{R_a+R_b}$$

I gave it a try:

$$ \begin{align} R & = R_1||(R_2||R_3 +R_4||R_5)\\ & = R_1||(R_{23} + R_{45})\\ & = \frac{R_1(R_{23} + R_{45})}{R_1+R_{23} + R_{45}}\\ R_{23} & = \frac{R_2R_3}{R_2+R_3} = \frac{20.46}{9.5}\\ R_{45} & = \frac{R_4R_5}{R_4+R_5} = \frac{56}{15.6}\\ R & = \frac{1(\frac{20.46}{9.5} + \frac{56}{15.6})}{1+\frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5}}{\frac{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{15.6 \times 9.5}} =\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5} \\ R & \approx 0.85 \end{align} $$

voltage \$U_{ab}\$ via voltage divider

I use U for voltage, not V.

\$R_1\$ being in parallel to the rest of the resistors means that there's the same voltage over both of them. The voltage divider divides the voltage \$U_{ac}\$ into \$U_{ab} + U_{ac}\$

The general rule for a voltage divider for two resistors in \$R_{ab}\$ and \$R_{bc}\$ in row: $$\frac{U_{ab}}{U_{ac}}=\frac{R_{ab}}{R_{ac}}=\frac{R_{ab}}{R_{ab} + R_{bc}}$$

I gave it a try: $$ \begin{align} \frac{U_{ab}}{U_{ac}} &= \frac{R_{23}}{R_{23} + R_{45}} \\ &= \frac{ \frac{20.46}{9.5}}{ \frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{ \frac{20.46}{9.5}}{ \frac{20.46 \times 15.6 + 56 \times 9.5}{9.5 \times 15.6}} = \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5}\\ &\approx 0.3750 \\ U_{ac} &= 5 \\ U_{ab} &= \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 1.8750 \end{align} $$

current \$I_{R_2}\$ via current divider

The current divider divides the current \$I_{ab}\$ into \$I_{R_2} + I_{R_3}\$

The general rule for a current divider for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel: $$\frac{I_{a}}{I_{a} + I_{b}}=\frac{I_{a}}{I_{ab}}=\frac{R_{a} || R_{b}}{R_{a}}= \frac{R_aR_b}{R_a(R_a +R_b)}=\frac{R_{b}}{R_{a} + R_{b}}$$

The general rule for resistance, voltage and current (ohm's law) : $$R = \frac{U}{I} \iff I = \frac{U}{R}$$

I gave it a try:

$$ \begin{align} I_{ab} &= \frac{U_{ab}}{R_{ab}} = \frac{U_{ab}}{R_{23}}\\ &= \frac{\frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5}{\frac{20.46}{9.5}} = \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.8706\\ \frac{I_{R_2}}{I_{ab}} &= \frac{R_3}{R_2 + R_3} = \frac{3.3}{9.5} \\ &\approx 3.4747 \\ I_{R_2} &= \frac{R_3}{R_2 + R_3} \times I_{ab} = \frac{3.3}{9.5} \times \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5 = \frac{15.6 \times 3.3}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.3024\\ \end{align} $$

power \$P_{R_1}\$

Given the overall voltage \$U_{ac}\$ and the resistance \$R_1\$ the power \$P_{R_1}\$ can be calculated.

The general rule for power: $$P = U \times I$$ With ohm's law: $$P =\frac{U^2}{R}$$

I gave it a try: $$ \begin{align} P_{R_1} &=\frac{U_{ac}^2}{R} =\frac{5^2}{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}} =\frac{25 \times 15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{20.46 \times 15.6 + 56 \times 9.5} \\ &\approx 5.3528 \end{align} $$

Consider the equation that uses the reverse saturation current

$$I_C=I_S\cdot e^\frac{V_{BE}}{V_T}$$

And then consider what happens for a slight difference in Vbe: -

• If Vbe = 0.70 volts then Ic = 0.394 mA
• If Vbe = 0.68 volts then Ic = 0.183 mA
• If Vbe = 0.72 volts then Ic = 0.851 mA

So choose Vbe carefully or you'll be a mile out.

If you chose Vbe to be 0.72597 volts, Ic would equal 1.07 mA and match your 2nd derivation. It's easy to get fixated on what you believe are correct numbers.

As for the 2nd derivation using Kirchoffs, there is an omission that is relevant and will slightly lower the collector current. The omission I refer to is called \$r_E\$ and is the internal emitter resistance.

In simple terms it equals \$\dfrac{V_T}{I_C}\$

Or about 26 ohms for an \$I_C\$ of about 1 mA and is fairly significant when you consider that \$R_E\$ is only 400 ohms. So, it's best not to get too fixated on some formulae. I'd be interested to know what a sim produces.