This community is so helpful, my thanks go out to any comment I got with helpful advice.

Here's another homework problem.

I've tried working most of it out, but I have a feeling that I'll have to redo some of the calculations. I've never seen resistors in this type of network before. All I know is that it might be called a bridge circuit?

I just need help finding: The total resistance of the circuit, finding Va using the voltage divider rule, finding V1 and V2, and calculating I1 with a direction.

Here are my calculations:

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Original circuit

enter image description here


You must apply the rule for determining the aggregate value of the parallel resistances before applying the voltage divider rule. Redraw the circuit like this and it may help you see what you need to do better:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thanks for showing me how to draw this schematic, but isn't a wire connected between R1 and R4 and also connect across between R3 and R5. \$\endgroup\$ – Omuse May 23 '15 at 13:35
  • \$\begingroup\$ feel free to correct me if I'm wrong. \$\endgroup\$ – Omuse May 23 '15 at 13:35
  • \$\begingroup\$ No, the wire does not connect at the point between R1 and R4. The arch in the line on the original schematic means skip this point. \$\endgroup\$ – JamieSee May 24 '15 at 16:32

As you have already obtained, the given circuit can be redrawn by combining resistors as shown below.


simulate this circuit – Schematic created using CircuitLab

Now the voltage \$V_a\$ will be given by the voltage divider made of \$R_x\$ and \$R_y\$. Not by that made of \$R_3\$ and \$R_5\$ as you have obtained. ie,

$$V_a = \frac{R_y}{R_x+R_y}\times 32V$$

And once you obtain \$V_a, V_1\$and \$V_2\$ can be calculated easily because \$V_2 = V_a\$ and \$V_1+V_2=32V\$. Since you have already calculated total resistance \$R_T\$, calculating \$I\$ is straightforward.

  • 1
    \$\begingroup\$ $$V_\mathrm{a} = \frac{V_{s} \times R_{y}} {R_{x} + R_{y}} \mathrm {\ watts}$$ \$\endgroup\$ – EM Fields May 22 '15 at 19:55
  • \$\begingroup\$ @EMFields thank you for pointing out. but why did you write \$\texttt{watts}\$? \$\endgroup\$ – nidhin May 23 '15 at 5:12
  • \$\begingroup\$ Oops... Sorry, I wasn't paying attention; it should read: $$V_a = \frac{V_s \times R_y}{R_x+R_y} \ volts$$ \$\endgroup\$ – EM Fields May 23 '15 at 7:50
  • \$\begingroup\$ Thanks for clearing that up so the voltage divider rule can still be used without Rz disrupting the calculations for Va? \$\endgroup\$ – Omuse May 23 '15 at 13:40
  • \$\begingroup\$ @Omuse Yes. The voltage across \$R_x+R_y\$ will be divided across these resistors as given by voltage divider formula. \$R_z\$ will not disrupt the calculation. \$\endgroup\$ – nidhin May 23 '15 at 14:31

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