An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
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Answer:
Vab ≈ 3.426 V
Explanation:
First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...
R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600
Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:
Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1
The voltage at Vb is also a divider from V1:
Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1
The parallel branches containing Va and Vb have an effective resistance of ...
(1030)(1710)/(1030+1710) = 642.81
That forms a divider with R1 to give V1:
V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V
The difference Va-Vb is ...
Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V
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We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...
(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0
(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0
(V1 -Va)/R3 -Va/R4 = 0
The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.
