# How to solve this resistive network using divider rules and ohms law

I have a homework problem I've been trying to solve. So far I think I'm using the right formulas for the calculations.

I just need help finding: The total resistance of the circuit, finding Vab using the voltage divider rule, finding IR2 using the current divider rule, and finally finding the power dissipated by R1.

Here's what i've worked out so far.

This is the original circuit

Maybe I need to calculate the resistor in parallel with the battery, and add it as the total resistance. But how would you calculate that, and would it work?

I re-did (R2//R3) in series with (R4//R5). And calculated the total resistance.

I tried calculating Vab but I'm not sure if I'm correct. Here's my calculations.

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}

• Cool, so this confirms my calculations for total resistance. but your method is faster. I should try this way out next time. – Omuse May 22 '15 at 15:58
• @Omuse I added the voltage, current and power with all the g(l)ory details. Take the results with a grain of salt and feel free to edit this post if you find a mistake. – Magic Smoke May 22 '15 at 17:43
• Thanks I think everything looks correct. I'll check when I my homeworked is checked. – Omuse May 23 '15 at 13:24

You are very close but you made a mistake that is keeping you from completion.

R2 is in parallel with R3 to equal R2||R3.

R4 is in parallel with R5 to equal R4||R5.

Next note that R2||R3 is in series with R4||R5 so these add together as: R2||R3 + R4||R5 to equal RRight.

Finally RRight is in parallel with R1.

I think you can finish from there.

Edit Actually if you are trying to find Vab then use R2||R3 as the top resistor in a voltage divider with R4||R5 across 5V. R1 need not be considered unless you were trying to compute total power consumed from the 5V supply.

• I calculated my total resistance to be 0.85 kilo ohms. – Omuse May 22 '15 at 12:19
• So just to be clear, I should use R2||R3 as the top resistor and R4||R5 as the bottom resistor and 5v as the power supply passing through them? – Omuse May 22 '15 at 12:24
• Sorry I'm a little confused. Vab means the voltage across R2||R3 right? – Omuse May 22 '15 at 12:25
• I re-edited my post again I tried working out Vab. But I feel like I've made a mistake. – Omuse May 22 '15 at 12:50

Here's an easy way to go about it:

Since R1 and Rx are in parallel, their total resistance in the circuit would be:

$$Rt = \frac{R1 \times Rx}{R1+Rx} = \frac{1000 \Omega \times 5740 \Omega}{1000 \Omega + 5740 \Omega} = 851.6 \Omega$$

and the power consumed by R1 would be:

$$P = \frac{V_{S}{}^2}{R1}\ watts$$

• Thanks, so looking at this method, total current is the sums of all the current passing through each component in the circuit? – Omuse May 22 '15 at 15:56
• @Omuse: 1.) Yes, with one caveat, which is that since in a series circuit only one current flows through all of the elements in that series string, when several of those strings are paralleled, the current out of the source will be the sum of the string currents, not the sum of the individual component currents. 2.) You're welcome. :-) – EM Fields May 22 '15 at 16:15
• Ah I see, and each strings(nodes) will share the same current right? :D – Omuse May 22 '15 at 16:35
• I don't understand what you're asking. – EM Fields May 22 '15 at 16:39
• Sorry, What I mean is, will the current still remain the same if it passing through one wire that later branches off into two? – Omuse May 22 '15 at 16:49