A circuit is constructed with five resistors and a battery as shown. The battery voltage is V = 12 V. The values for the resistors are: R1 = 78 Ω, R2 = 150 Ω, R3 = 132 Ω, and R4 = 83 Ω. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.

I have a circuit constructed in the figure below (fig 1)

fig 1

I've been able to solve the following, based on the battery being 12[v]

R, Ohm, I

R1, 78, 0.15384

R2, 150, 0.08

R3, 132, 0.0909

R4, 83, 0.1446

R5, 0*, 0.0

R5: $V = iR --> 12 = (0)R --> 12 != 0$

Meaning that no current or voltage pass through R5.


Here's my question:

How do I solve for the magnitude of voltage at R2?


I was able to solve for V @ R1 by:

$$ R_{1}(i) = \frac{V}{R_{1} + R_{3}} = 0.057 [A] $$

I understand that I need to break the circuit into "systems". I know that to compute for V of R2 I need to consider V of R5, but I do not know how to calculate V or R for R5, since I'm not really sure how to handle the bridge at R4 and the fact that it's resistance has to be zero (since i=0).

My best guess so far is this:

$$ V_{2} = \frac{V R_{2}}{R_{2} X} $$

Where $X$ is a sum of R2 and R5


Any help is appreciated.

  • $\begingroup$ So after some insight from akhmeteli and Alfred Centauri, it made sense to just assume that $R_{X}$ is in series with $R_{2}$ and not some funny loop with any others. $$ V_{Total} = (R_{1} + R_{3})i_{1} = (R_{2} + R_{X})I_{2} \\ R_{X} = 253.83 [V] $$ $\endgroup$ – Adam A. Sep 30 '13 at 20:49
  • $\begingroup$ @AdamA., under the condition that I4 = 0 (and only under that condition), R4 can actually be removed from the circuit without changing any voltage or current in the circuit so your approach above is valid. But, there is an easier method still. In order for it to be the case that I4 = 0, the bridge must be balanced (the voltage across R4 must be zero) which implies that $R_X = R_3 \dfrac{R_2}{R_1}$. That's how simple it is to find the value of Rx. $\endgroup$ – Alfred Centauri Oct 1 '13 at 12:59

I don't understand what you were doing, but I would do the following. As $I_4=0$, the same current $I_1$ flows through $R_1$ and $R_3$, and the same current $I_2$ flows through $R_2$ and $R_x$ (that means, by the way, that you miscalculated the currents). So $(R_1+R_3)I_1=(R_2+ R_x)I_2$=12V and $R_1 I_1=R_2 I_2$, as the voltage difference on $R_4$ vanishes. So you have three equations and three unknowns.


... the current that flows through resistor R4, is zero. How do I solve for the magnitude of voltage at R2?

The crucial insight here is that, since $I_4 = 0$, the voltage across $R_4$, by Ohm's Law, must be zero.

Thus, the voltage across $R_2$ must equal the voltage across $R_1$ (Use KVL to convince yourself of this if it isn't immediately apparent).

But, and once again, since $I_4 = 0$, $R_1$ and $R_3$ are in series (the current through $R_1$ is through $R_3$) and thus, the voltage across $R_1$ is given by voltage division:

$$V_{R_1} = 12V \dfrac{R_1}{R_1 + R_3} = V_{R_2}$$


Why is it that nobody recognizes this problem as simply a balanced Wheatstone bridge ?

If the current in R4 is zero the bridge is balanced and the value of either the Voltage or R4 is totally irrelevant.

R.y /R3 = R2/R1

Since this is a homework or hardware problem I can't give you the answer; also we are not allowed to bother people with equations, so I didn't give you the exactly correct equation either, so you will have to figure out your own equation.

  • $\begingroup$ The op plainly asks How do I solve for the magnitude of voltage at R2? so, contrary to your assertion, the voltage is relevant. $\endgroup$ – Alfred Centauri Sep 30 '13 at 19:45
  • $\begingroup$ The Battery Voltage IS irrelevant. The "OP said he COULD solve for the Voltage at R1. So the answer to his question is "Same Voltage as at R1" for all values of R4 and all Battery Voltages. $\endgroup$ – user26165 Sep 30 '13 at 19:53
  • $\begingroup$ I'm not looking for a solution to this assignment, I'm asking for some insight on how to handle the bridge. When I've tried to compute using the bridge and without I can not seem to get a value for $R_{x}$ that makes any sense. I guess the underlying question here is, do I handle the bridge as a series or a parallel circuit? $\endgroup$ – Adam A. Sep 30 '13 at 20:15
  • $\begingroup$ @GeorgeE.Smith, thus the voltage is relevant since the voltage across R1 depends on it. You've contradicted yourself. $\endgroup$ – Alfred Centauri Sep 30 '13 at 20:23
  • $\begingroup$ @AdamA., the question of finding Rx is separate from finding the voltage across R2. If all you want is Rx, you making the problem way to hard. $\endgroup$ – Alfred Centauri Sep 30 '13 at 20:25

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