Nice and tricky question on electric circuits

[mmh37]

Hi, this is a very nice but (at least for me) quite confusing problem on electric circuits:

Before you read this, it will be helpful to have had a look at the attached picture (sorry - the quality is quite nasty)

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By considering each half of the circuit on the left below as a potential divider, one can show that

Z1/Z2 = Z3/Z4

The bridge circuit on the right of the picture is said to be balanced when the detector D registers no voltage difference between its terminals. Use the above equation to find formulae for R and L in terms of the other components when the circuit is balanced.

OK, so this is what I tried:

Z1= R + XL

Z2 = R2

Z3 = R3

[tex]Z4 = (\frac {1} {R4} + iwC4)^{-1} [/tex]

Equation 1

as derived from Z1/Z2 = Z3/Z4

therefore

[tex] R + iwL = R2*R3*(\frac {1} {R4} + iwC4) [/tex]

Equation 2

Now, regard the series connection on the respective sides of the potential divider.

given: U(Z2) = U(Z4) (A)


left hand side:

[tex] U(left) = ( Z1 + Z2)*I = \frac {U(0)} {2} [/tex]

solve for I to calculate

[tex] U (Z2) = \frac {Z(2)*U(0)} {2* (Z3 + Z4)} [/tex]

right hand side:

like lhs

[tex] U(Z4) = \frac {Z(4)*U(i)} {2(Z(1)*Z(2)}[/tex]


So now we put that in eq. (A)

to get:

[tex] R + iwL = R3/R2*(\frac {1} {R4} + iwC4)^{-2} [/tex]

Cool,

But now I don't know how to solve for L and R as w is not given and I don't know how to deal with those complex numbers to find L and R.

Can anyone help?? That would be absoluetly awesome!

 

[Jhero]

Hello!

I can definitely understand how this problem can be confusing. But don't worry, we can break it down step by step to solve it. First of all, let's look at the potential divider equation, Z1/Z2 = Z3/Z4. This equation is based on the principle of Kirchhoff's voltage law, which states that the sum of voltages around a closed loop in a circuit must equal zero. In this case, we have two loops - one on the left side of the circuit and one on the right side. By applying Kirchhoff's law to each loop, we can come up with the potential divider equation.

Now, let's move on to the bridge circuit on the right side of the picture. When the circuit is balanced, the detector D registers no voltage difference between its terminals. This means that the voltage across Z1 must be equal to the voltage across Z3, and the same goes for Z2 and Z4. Using this information, we can write the following equations:

V1 = V3 (equation 1)
V2 = V4 (equation 2)

Where V1 and V2 are the voltages across Z1 and Z2, and V3 and V4 are the voltages across Z3 and Z4, respectively.

Now, let's substitute the equations for Z1, Z2, Z3, and Z4 into equations 1 and 2. This will give us:

R + iωL = R2 (equation 3)
R3 = \frac {1} {R4} + iωC4 (equation 4)

Next, we can rearrange equation 4 to solve for R4:

R4 = \frac {1} {R3 - iωC4}

Now, we can substitute this into equation 3 to get:

R + iωL = R2 * (\frac {1} {R3 - iωC4})

We can then expand the right side of the equation to get:

R + iωL = \frac {R2} {R3 - iωC4}

Now, we can rearrange this to solve for L:

L = \frac {R2} {R3*R - iω(R3*R2 + C4*R2)}

And we can also solve for R:

R = \frac {R2} {R3 - i