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# How to measure medium resistances? (Resistance Measurement Methods)

The resistances that range from about **1Ω**
**to about 100 kΩ **are classified as **medium resistances**. The resistances of most of electrical apparatus are the examples of medium resistances.

## Measurement of Medium Resistances

To measure the medium resistances following methods are used −

Ammeter-Voltmeter Method

Substitution Method

Wheatstone Bridge

Carey-Foster Slide-Wire Bridge Method.

### Ammeter-Voltmeter Method

In this method, current through the unknown resistor (R_{x}) and the potential drop across it are simultaneously measured. The readings are obtained by ammeter and voltmeters respectively. There are two ways in ammeter and voltmeters may be connected for measurement as,

**Case 1 **– When voltmeter is directly connected across the resistor, then the ammeter measures current flowing through the unknown resistance (R_{x}) and the voltmeter.

Current through ammeter = Current through(𝑅_{x}) + Current through voltmeter

$$\mathrm{I=I_{R_{x}}+I_{V}}$$

$$\mathrm{\Rightarrow\:I_{R_{x}}=I-I_{V}}$$

Therefore, the value of unknown resistance,

$$\mathrm{R_{X}=\frac{V}{I_{X}}=\frac{V}{I-I_{V}}=\frac{V}{I-(V/R_{V})}\:\:\:...(1)}$$

**Case 2** – When the ammeter is connected such that it measures only the current flowing through the unknown resistor (R_{x}), then the voltmeter measures voltage drop across the ammeter and R_{x}.

Therefore,

$$\mathrm{V=IR_{A}+IR_{X}=I(R_{A}+R_{X})}$$

$$\mathrm{\Rightarrow\:R_{X}=\frac{V}{I}-R_{A}\:\:\:\:...(2)}$$

### Substitution Method

**Step 1** – In this method, first the unknown resistance (R_{x}) is put into the circuit and note the value of current.

**Step 2 **– Then the resistance R_{x} is removed and it is *substituted* by a known variable resistance R which is varied so that the value of current is same in both the cases. This value of R is equal to the value of unknown resistance.

### Wheatstone Bridge

The *Wheatstone bridge* method is the most accurate method for the measurement of resistances.

The bridge consists of four resistive arms, source of emf and a galvanometer (null detector). The current through the galvanometer depends upon the potential difference between the points B and D. The bridge is said to be *balanced* when the potential difference across the galvanometer is zero so that there is no current flows through the galvanometer.

For the *balanced Wheatstone bridge*,

$$\mathrm{PS=QR_{x}}$$

$$\mathrm{\Rightarrow\:R_{x}=\frac{PS}{Q}\:\:\:...(3)}$$

### Carey-Foster Slide-Wire Bridge Method

The circuit of this bridge is the elaborated form of Wheatstone bridge and is specially used for comparing two nearly equal resistances.

The circuit consists of four resistive arms, where R_{1} and R_{2} are the ratio arms, R_{3} is standard resistance and R_{x} is unknown resistance. A slide-wire (m-n) of length l and uniform crosssection is connected between R_{3} and R_{x}. The slide wire has a resistance of r Ω per unit length.

The resistances R_{1} and R_{2} are adjusted so that the ratio $\mathrm{(\frac{R_{1}}{R_{2}})}$ is approximately equal to the ratio $\mathrm{(\frac{R_{x}}{R_{3}})}$. This balance is obtained by adjusting the sliding contact on the slide-wire. Then, For the *first balance*, let *l _{1}* is the distance of the sliding contact from the point m of the slide wire. Thus,

$$\mathrm{\frac{R_{1}}{R_{2}}=\frac{R_{x}+l_{1}r}{R_{3}+(l-l_{1})r}\:\:\:...(4)}$$

For the *second balance*, let *l _{2}* is the distance from the point m of the slide wire and the resistances R

_{x}and R

_{3}are interchanged, then

$$\mathrm{\frac{R_{1}}{R_{2}}=\frac{R_{3}+l_{2}r}{R_{x}+(l-l_{2})r}\:\:\:...(5)}$$

From eqns. (4) and (5), we get,

$$\mathrm{\frac{R_{x}+l_{1}r}{R_{3}+(l-l_{1})r}=\frac{R_{3}+l_{2}r}{R_{x}+(l-l_{2})r}}$$

Hence, the difference (R_{3} – R_{x}) is obtained from resistance of the slide wire between the two balance points.

Now, the value of r i.e. resistance per unit length of the slide wire is obtained by connecting a known high resistance in parallel with R_{3}, that reduces the effective value to R_{3}′, thus again repeat the above procedure to obtain new balance points 𝑙1′and 𝑙_{2}′, so that,

$$\mathrm{R_{3}^{'}-R_{x}^{'}=r(l_{1}^{'}-l_{2}^{'})\:\:\:\:...(7)}$$

Dividing eqn. (6) by eqn. (7), we obtain,

$$\mathrm{\frac{R_{3}-R_{x}}{R_{3}^{'}-R_{x}^{'}}=\frac{r(l_{1}-l_{2})}{r(l_{1}^{'}-l_{2}^{'})}}$$

$$\mathrm{\Rightarrow\:R_{x}=\frac{R_{3}(l_{1}^{'}-l_{2}^{'})-R_{3}^{'}(l_{1}-l_{2})}{(l_{1}^{'}-l_{2}^{'}-l_{1}+l_{2})}\:\:\:\:....(8)}$$

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