To Determine Resistance per CM of a Given Wire by Plotting a Graph for Potential Difference versus Current

To determine the resistance per centimetre of a wire, we can use Ohm's law, which states that the potential difference (voltage) across a conductor is directly proportional to the current flowing through it, provided the temperature remains constant. By measuring the potential difference and the corresponding current for different lengths of the wire, we can plot a graph of potential difference (V) versus current (I). The slope of this graph gives the resistance of the wire. Once we have the resistance, we can calculate the resistance per centimetre by dividing the total resistance by the length of the wire. This method provides a clear understanding of how resistance varies with length and helps in determining the precise resistance per unit length of the wire.

Aim

To determine resistance per cm of a given wire by plotting a graph of potential difference versus current.

Apparatus

A resistance wire, a voltmeter (0-3) V and an ammeter (0-3) A of appropriate range, a battery (battery eliminator), a rheostat, a meter scale, a one-way key, connecting wires and a piece of sandpaper.

Theory

According to Ohm's law the current flowing through a conductor is directly proportional to the potential difference across its ends provided the physical conditions (temperature, dimensions, pressure) of the conductor remain the same.

If I be the current flowing through a conductor and V be the potential difference across its ends, then according to, Ohm's Law,

$\begin{aligned} & I \propto V \\ & V \propto I \\ & \text { or } V=R I\end{aligned}$

where R is the constant of proportionality. It is known as the resistance of the conductor.

So we can use $R=\frac{V}{I}$

R depends upon the nature of the material, temperature, and dimensions of the conductor.

In S.I. units, the potential difference V is measured in volts and in the current in ampere, the resistance R is measured in ohm.

(1) To establish the current-voltage relationship, it is to be shown that the ratio $\frac{V}{I}$ remains constant for a given resistance, therefore a graph between the potential difference (V) and the current (I) must be a straight line.

(2) The constant ratio gives the unknown value of resistance, $\left(\frac{V}{I}=R\right)$

Circuit Diagram

Procedure

1. Arrange the apparatus in the same manner as given in the arrangement diagram.
2. Clean the ends of the connecting wires with sandpaper to remove the insulations, if any.
3. Make neat, clean and tight connections according to the circuit diagram. At the same time, making connections ensures that + ve marked terminals of the voltmeter and ammeter are joined towards the + ve terminal of the battery.
4. Determine the least count of the voltmeter and ammeter, and also note the zero error, if any.
5. Insert the key K, slide the rheostat contact and see that the ammeter and voltmeter are working properly.
6. Adjust the sliding contact of the rheostat such that a measurable current passes through the resistance coil or the resistance wire.
7. Note down the value of the potential difference $V$ " from the voltmeter and current I from the ammeter.
8. Shift the rheostat contact slightly so that both the ammeter and voltmeter show full division readings and not in the fraction.

9. Record the readings of the voltmeter and ammeter.

Note- In the case of battery eliminator, follow these steps:
Turn the knob at 2 V in a battery eliminator and put the constant point in the rheostat at the fixed position. Now record the reading in voltmeter and ammeter. Without disturbing the rheostat, turn the knob of the battery to a different voltage such that 4,6,8,10 and 12 Volts and record corresponding readings in the voltmeter and ammeter.
10. Take at least five sets of independent observations.
11. Cut the resistance wire at the points where it leaves the terminals, stretch it and find its length by the meter scale.
12. Record your observations.

Observations

1. Length

Length of the resistance wire $l=\ldots$
2. Range

Range of the given ammeter = $\qquad$
Range of the given voltmeter $=$ $\qquad$
3. Least count

Least count of ammeter = $\qquad$
Least count of voltmeter $=$ $\qquad$
4. Zero error

Zero error in ammeter, $\mathrm{e}_1=\ldots$.
Zero error in voltmeter, $e_2=\ldots$.

Calculations

1. Find the ratio of V and I for each set of observations.
2. Plot a graph between the potential difference V and current I, taking V along the X-axis and I along the Y-axis. The graph comes to be a straight line.

From the graph, the resistance can be calculated.
$
\begin{aligned}
& \text { in } \triangle A B C \quad \tan \theta=\frac{A B}{C B}=\frac{\Delta I}{\Delta V} \\
\Rightarrow & \cot \theta=\frac{\Delta V}{\Delta I}
\end{aligned}
$
But $R=\frac{\Delta V}{\Delta I}$
So $R=\cot \theta$

$
R=\ldots \ldots \Omega
$

3. Constant ratio $\frac{V}{I}$ gives resistance of the wire.
4. Resistance of the wire per $\mathrm{cm}=\ldots$.

Result

1. Resistance per cm of the wire is ......
2. The graph between V and I is a straight line.

Solved Examples Based on Determine Resistance Per cm of a Given Wire by Plotting a Graph of Potential Difference Versus Current

Example 1: You have a wire with an unknown resistance, and you want to determine its resistance using Ohm’s law. You’re provided with the following information:
Voltage across the wire (V ) = 12 volts
Current passing through the wire (I) = 2.5 amperes
Using Ohm’s law (V = I · R), calculate the resistance of the given wire.

1) 4.8 ohms

2)3.4 ohms

3)2.6 ohms

4)5.9 ohms

Solution:

Use Ohm's law to calculate the resistance (R):

$
V=I \cdot R
$
Solve for R:

$
R=\frac{V}{I}=\frac{12}{2.5}=4.8 \circ \mathrm{hms} \ldots \ldots
$

Therefore, the resistance of the given wire is 4.8 ohms.

Hence, the answer is the option (1).

Example 2: A complex circuit consists of resistors R₁, R₂, and R₃ connected in series to a constant voltage source of 24V. The currents flowing through each resistor are measured and recorded:

Resistor Current (A) R₁ 0.5 R₂ 0.8 R₃ 1.2 Using this data, calculate the equivalent resistance of the entire circuit.

1) $8.5 \Omega$
2) $98.5 \Omega$
3) $59.5 \Omega$
4) $7.5 \Omega$

Solution:

In a series circuit, the total resistance (R_eq) is the sum of the individual resistances:

R_eq = R₁ + R₂ + R₃

Let’s use the provided data points to calculate the equivalent resistance:

Resistor Current (A) R₁ 0.5 R₂ 0.8 R₃ 1.2

We'll use Ohm's law to find the resistance of each resistor:

$
R=\frac{V}{I}
$

where V is the constant voltage $(24 \mathrm{~V})$ and I is the current.

$
\begin{aligned}
& R_1=\frac{24}{0.5}=48 \Omega \\
& R_2=\frac{24}{0.8}=30 \Omega \\
& R_3=\frac{24}{1.2}=20 \Omega
\end{aligned}
$
Now, calculate the equivalent resistance:

$
R_{e q}=R_1+R_2+R_3=48+30+20=98 \Omega
$

Hence, the equivalent resistance of the entire circuit is $98 \Omega$.

Hence, the answer is the option (2).

Example 3: A parallel circuit consists of three resistors R₁, R₂, and R₃ connected in parallel to a constant voltage source of 36V. The currents flowing through each resistor are measured and recorded:

Resistor Current (A) R₁ 1.5 R₂ 2.0 R₃ 2.5 Using this data, calculate the equivalent resistance of the entire parallel circuit.

1) $88 \Omega$
2) $66 \Omega$
3) $24 \Omega$
4) $50 \Omega$

Solution:

In a parallel circuit, the reciprocal of the total resistance (1/ R_eq) is the sum of the reciprocals of the individual resistances:

$\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

Let’s use the provided data points to calculate the equivalent resistance:

Resistor Current (A) R₁ 1.5 R₂ 2.0 R₃ 2.5

We'll use Ohm's law to find the resistance of each resistor:

$
R=\frac{V}{I}
$

where V is the constant voltage $(36 \mathrm{~V})$ and I is the current.

$
\begin{aligned}
& R_1=\frac{36}{1.5}=24 \Omega \\
& R_2=\frac{36}{2.0}=18 \Omega \\
& R_3=\frac{36}{2.5}=14.4 \Omega
\end{aligned}
$
Now, calculate the equivalent resistance:

$
\frac{1}{R_{e q}}=\frac{1}{24}+\frac{1}{18}+\frac{1}{14.4}=0.0417 \Omega
$
Taking the reciprocal:

$
R_{e q}=\frac{1}{0.0417}=24 \Omega
$

Hence, the equivalent resistance of the entire parallel circuit is $24 \Omega$.

Hence, the answer is the option (3).

Example 4: A wire is connected to a voltage source, and a current of 0.5 amperes flows through the wire. The voltage across the wire is measured to be 12 volts. Determine the resistance (R) of the given wire using Ohm’s law.

1)160ohms

2)170 ohms

3)200 ohms

4)24 ohms

Solution:

Ohm’s law states that the resistance (R) of a circuit element can be calculated using the equation $R=\frac{V}{I}$ where V is the voltage across the element and I is the current flowing through it.

Given the voltage (V ) as 12 volts and the current (I) as 0.5 amperes, we can substitute these values into Ohm’s law to find the resistance (R):

$R=\frac{12 \text { volts }}{0.5 \text { amperes }}=24 \mathrm{ohms}$

Hence, the resistance (R) of the given wire is 24 ohms.

Hence, the answer is the option (4).

Summary

To determine the resistance per centimetre of a wire, we measure the potential difference and current for different wire lengths and plot a graph of voltage versus current. The slope of this graph gives the wire's resistance, which, when divided by the length, gives the resistance per centimetre. This method helps us understand how resistance changes with the wire's length.