How To Find Internal Resistance Of A Cell Using Potentiometer

Usually while studying the theory behind the electrical circuits it is assumed that all the components of the circuits are platonic. In an ideal state, the battery of the system is causeless to take zero resistance. But in reality, information technology is non the case. While working with batteries in real life, it s noticed that they accept some internal resistance which affects the current in the circuit. Sometimes it is not specified on the battery and it changes as the battery discharges. In that instance, it becomes essential to calculate the internal resistance of the battery being used. Allow's await at how it's done in detail.

Internal Resistance of a Bombardment

It is known that a large emf-based battery has more size than the batteries with less emf. These batteries comprise more energy and thus can deliver larger currents. Discover that a 12V bombardment of a truck can deliver more current than a 12V bombardment present in a motorbike. The reason behind this can exist attributed to the fact that the battery of the truck has less internal resistance than the battery of a motorbike.

Internal resistance is the inherent resistance that is present inside a voltage source.

The figure above shows ii fundamental parts of a voltage source. The emf present inside the battery and the resistance. This emf is denoted by E while the internal resistance is denoted past r, both of them are series. The smaller the internal resistance for the battery, the more than electric current it is able to supply to the excursion. The internal resistance of a battery tin can deport in complex means, equally the bombardment depletes the internal resistance of the bombardment increases. Just it may also depend on the magnitude and the direction of the electric current through a voltage source, its temperature, and fifty-fifty the cloth the bombardment is fabricated up of.

Potentiometer

Potential is used to measure out the potential departure between 2 points. Information technology is sometimes also used to compare the eastward.grand.f of two cells or measure the internal resistance of a battery. At a structural level, it is a device that consists of a long wire of uniform cantankerous-sectional area and of 10m in length. While using a potentiometer, one should make certain that the wire that is begin used in the device should have a uniform cantankerous-sectional surface area and low resistivity, and loftier-temperature coefficient. The wires are stretched parallel to each other and are joined in series through the copper strips. A meter scale is also attached to the wooden board.

Relation betwixt E.K.F and Potential Difference Across a Cell

Allow'due south consider a cell with eastward.m.f "Eastward" and internal resistance "r". The jail cell is continued to an external resistance of "R". In that example, the total resistance of the circuit becomes R + r. The current I in the circuit now will exist given by,

I = East/(R+ r)

⇒ E = I (R + r)

Thus, the potential across the external resistance,

Five = IR = E – Ir

This equation shows that V is less than the due east.m.f of the prison cell. This fall in the potential is due to the potential drop in the internal resistance of the battery. The equation given above tin can be rearranged as,

$\frac{r}{R} = \frac{E - V}{V}$

The internal resistance will exist given past,

$r = R\frac{E - V}{V}$

Rheostat tin be adapted on the potentiometer to obtain the balancing lengths l₁ and l_twofor open and closed circuits respectively. The figure shows this setting,

Allow "k" be the potential gradient on the wire on the potentiometer.

E = kl₁ and V = kl₂

The above equation for the internal resistance tin exist modified by,

$r = R\frac{kl_1 - kl_2}{kl_1}$

⇒ $r = R\frac{l_1 - l_2}{l_1}$

Sample Problems

Question ane: Find the current that will menses within the battery of 10 Volts and 0.04 ohms internal resistances in case its terminals are connected with each other.

Answer:

The current in that case will be given by uncomplicated application of ohm'due south law.

V = 10V

r = 0.04 ohms.

5 = IR

Plugging the values in the equation,

I = V/R

⇒ I = ten/0.04

⇒ I = 250 A

Question 2: Find the current that volition flow inside the bombardment of 40 Volts and 10 ohms internal resistances in example its terminals are continued with each other. Observe the terminal voltage of the battery.

Reply:

The current in that case volition be given by unproblematic application of ohm's law.

V = 40 V

R= 10 ohms.

V = IR

Plugging the values in the equation,

I = V/R

⇒ I = 40/10

⇒ I = 4 A

The terminal voltage of the bombardment is given past,

V = emf – Ir

Given , emf = twoscore V, I = 4A and r = 10

V = emf – Ir

⇒ V = 40 – (four)(10)

⇒ 5 = 0 V

Question 3: Find the electric current that will menstruation within the battery of 20 Volts and 4 ohms internal resistances and vi ohms load resistance in series. Find the last voltage of the battery.

Answer:

The current in that case will be given by simple application of ohm'due south law.

I = $\frac{\text{emf}}{R_{\text{load}} + r}$

emf = 20 V

R_load= 6 ohms.

r = iv

plugging the values in the equation,

I = $\frac{\text{emf}}{R_{\text{load}} + r}$

⇒ I = $\frac{20}{4 + 6}$

⇒ I = two A

The terminal voltage of the battery is given past,

V = emf – Ir

Given, emf = 20V, I = 2A and r = 4

V = emf – Ir

⇒ V = 20 – (2)(four)

⇒ Five = 20 – eight

⇒ V = 12V

Question 4: Discover the internal resistance of the battery if the potentiometer is balanced at lengths 50₂ = 120cm and l₁ = 300 cm from ane stop. Assume the external resistance is 10 Ohms.

Answer:

The internal resistance when measured through potentiometer is given by,

$r = R\frac{l_1 - l_2}{l_1}$

Given:

l₁ = 300 cm

l₂ = 120 cm

R = 10

Plugging the values in the equation,

$r = R\frac{l_1 - l_2}{l_1}$

⇒ $r = (10)(\frac{300 - 120}{300})$

⇒ $r = (10)(\frac{180}{300})$

⇒ r = half-dozen Ohms

Question 5: Find the internal resistance of the battery if the potentiometer is counterbalanced at lengths l_ii = 60cm and l₁ = 150 cm from one stop. Assume the external resistance is 20 Ohms.

Answer:

The internal resistance when measured through potentiometer is given by,

$r = R\frac{l_1 - l_2}{l_1}$

Given:

l_one = 60 cm

l_ii = 150 cm

R = twenty

Plugging the values in the equation,

$r = R\frac{l_1 - l_2}{l_1}$

⇒ $r = (20)(\frac{150 - 60}{150})$

⇒ $r = (20)(\frac{90}{150})$

⇒ r = half-dozen Ohms