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**1.
****DC
BRIDGE NETWORKS**

The DC bridge is an electrical circuit for the precise measurement of resistances. The best known bridge circuit is the Wheatstone bridge, named after Sir Charles Wheatstone (1802 â€“ 1875), an English physicist and inventor.

The Wheatstone bridge circuit is shown in the figure below. The interesting feature of this circuit is that if the proyducts of the opposite resistances (R1R4 and R2R3) are equal, the current and voltage of the middle branch is zero, and we say that the bridge is balanced. If three of the four resistors (R1, R2, R3, R4) are known, we can determine the resistance of the fourth resistor. In practice the three calibrated resistors are adjusted until the voltmeter or ammeter in the middle branch reads zero.

**Wheatstone
bridges**

Letâ€™s prove the condition of balance.

When in balance, the voltages on R1 and R3 must be equal:

therefore

R_{1
}R_{3}+R_{1} R_{4} = R_{1} R_{3} +
R_{2} R_{3}

Since
the term R_{1} R_{3} appears on both sides of the equation, it
can be subtracted and we get the condition of balance:

R_{1}
R_{4} = R_{2} R_{3
}

In TINA you can simulate balancing the bridge by assigning hotkeys to the components to be changed. To do this, double click on a component and assign a hotkey. Use a function key with the arrows or a capital letter, e.g. A to increase and another letter, e.g. S to decrease the value and an increment of say 1. Now when the program is in interactive mode, (the DC button is pressed) you can change the values of the components with their corresponding hotkeys. You can also double-click on any component and use the arrows on the right side of the dialog below to change the value.

**Example
**

Find
the value of R_{x} if the Wheatstone-bridge is balanced. R_{1}
= 5 ohm, R_{2} = 8 ohm,

R_{3}
= 10 ohm.

The
rule for R_{x}

Checking with TINA:

If you have loaded this circuit file, press the DC button and hit the A key a few times to balance the bridge and see the corresponding values.

**2.
****AC
BRIDGE NETWORKS
**

The same technique can also be used for AC circuits, simply by using impedances instead of resistances:

In this case, when

**Z**_{1
}**Z**_{4}
= **Z**_{2} **Z**_{3}

the bridge will be balanced.

If
the bridge is balanced and for example **Z**_{1, }**Z**_{2}
, **Z**_{3 }are known

**Z**_{4}
= **Z**_{2} **Z**_{3}
/** Z**_{1}

Using an AC bridge, you can measure not only impedance, but also resistance, capacitance, inductance, and even frequency.

Since
equations containing complex quantities mean two real equations (for the
absolute values and phases *or* real and imaginary
parts) balancing an AC circuit
normally needs two operating buttons but also two quantities can be
simultaneously found by balancing an AC bridge. Interestingly
the balance condition of many AC bridges are independent of the
frequency. In the following we will introduce the most well known bridges, each
named after their inventor(s).

**Schering
â€“ bridge: measuring capacitors with series loss.
**

Find C so that the ammeter reads zero in the Schering-bridge. f = 1 kHz.

The bridge will be balanced if:

**Z**_{1
}**Z**_{4}
= **Z**_{2} **Z**_{3}

In our case:

after multiplication:

The equation will be satisfied if both real and imaginary parts are equal.

In
our the bridge, only C and R_{x} are unknown. To find them we have to
change different elements of the bridge. The best solution is to change R_{4}
and C_{4} for fine-tuning, and R_{2} and C_{3} to set
the measurement range.

Numerically in our case:

independent of the frequency.

At the calculated values the current equals zero.

**Maxwell
bridge: measuring capacitors with parallel loss
**

Find
the value of the capacitor C_{1} and its parallel loss R_{1} if
the frequency f = 159 Hz.

The condition of balance:

**Z**_{1}**Z**_{4}
= **Z**_{2}**Z**_{3}

For this case:

The real and imaginary parts after multiplication:

R_{1}*R_{4}
+ j w
L_{1}*R_{1} = R_{2}*R_{3} + j w
R_{1} R_{2} R_{3}C_{1}

And from here the condition of balance:

Numerically
R_{1} = 10^{3}*10^{3}/10^{3} = 1 kohm,
C_{1} = 10^{-3}/10^{6} = 1 nF

In
the next figure you can see that with these value of C_{1} and R_{1}
the current really is zero.

**Hay bridge:
measuring inductances with series loss
**

Measure
the inductance L_{1} with series loss R_{4}.

The bridge is balanced if

**Z**_{1}**Z**_{4}
= **Z**_{2}**Z**_{3}

After multiplying, the real and imaginary parts are:

Solve
the second equation for R_{4}, substitute it into the first criteria,
solve for L_{1}, and substitute it into the expression for R_{4}:

These criteria are frequency dependent; they are valid only for one frequency!

Numerically:

Using
the interpreter:
om:=Vsw L:=C1*R2*R3 / (1+om*om*C1*C1*R1*R1) R:=om*om*R1*R2*R3*C1*C1 / (1+om*om*C1*C1*R1*R1) L=[5.94070853] R=[59.2914717] |

Checking the result with TINA:

**Wien-Robinson
bridge: measuring frequency
**

How can you measure frequency with a bridge?

Find the conditions for balance in the Wien-Robinson bridge.

The
bridge is balanced if R_{4}
ּ(R_{1} + 1/ **j **w
C_{1} ) = R_{2} ּR_{3} / (1 + **j**
w
C_{3} R_{3})

After multiplication and from the requirement of equality of the real and imaginary parts:

If C_{1} = C_{3} = C and R_{1}
= R_{3} = R the bridge will
be balanced if R_{2} = 2R_{4}
and the angular frequency:

`

Checking the result with TINA:

BODE PLOTS | RESONANT CIRCUITS |