Internal resistance is the opposition provided to the flow of current by the cells and the batteries. It results in the production of heat. **Ohms** is a unit to measure the internal resistance.

There are various formulas to determine internal resistance. We can find answers to any question if w are provided with the data. For example, to find Internal resistance we use this formula:

**e = I (r + R)**

In this formula, e is the EMF or electromotive force that is measured in ohms, I is the Currents which is measured in Amperes (A) and R is load resistance while **r is Internal resistance. **Ohms is the unit of measurement for internal resistance.

The formula provided previously is re-arranged in this form,

- e = Ir+ IR
- e = V + Ir

V is denoted as the Potential difference applied across the cell and I is representing the current flowing across the cell.

Note:The Electromotive force (emf) is always greater than the potential difference (V) of the cell.

Thus, knowing some of the parameters lead us to find others. I will b addressing many practice problems in this article, that will help you know the use of Physics in our daily life, and the ways to calculate the parameters along with formulas and descriptions. Just stick with me till the end.

On an open circuit, the potential difference between the battery terminals is 2.2 volts. The potential difference is reduced to 1.8 volts when it is connected across a resistance of 5 ohms. What exactly is internal resistance?

This is an open circuit. The internal resistance of the battery has no voltage drop across it in an open circuit. When a closed circuit is formed, the current flows through the internal resistance, causing a voltage drop and lowering the voltage across the battery.

In this case, you must identify the internal resistance. You measure the voltage across the circuit as it opens and closes, as well as the load resistance. To solve this problem, first, we need to gather the data that is provided in the statement and then predict what needs to be calculated.

Data:Potential difference V = 2.2 Volts , Load resistance Resistance= 5 ohms, drop of potential difference is 1.8 volts,Find the internal resistance.

To find that, we need to solve the following steps.

First, **we need to find the load current as**,

I = V/R so, 1.8/5 = 0.36A

Then, **Find Voltage drop of battery internal resistance:**

2.2V-1.8V=0.4V

So, knowing the current and voltage of internal resistance :

R=V/I, 0.4/0.36 gives 1.1 ohms

Hence the internal resistance is 1.1 ohms.

## In an open circuit, the potential difference between the terminals of a cell is 2.2 volts. The terminal potential difference is 1.8 volts with a resistance of 5 ohms across the terminals of the cell. What will the cell’s internal resistance be?

This is a simple question about two resistors connected in series across a 2.2 V source, one of which is 5 ohms. So the question is, what is the other resistance in the series combination, internal battery resistance?

This is incredibly simple. First, Draw a 2.2 Volt cell, then an R (internal resistor), a 5-ohm external resistor, and finally return to the source.

**Across 5 ohms, there is a 1.8-volt drop.**

What exactly is the internal resistor if the current flowing through it is I = 1.8/5 amps = 0.36 A?

Let’s take a look at it,

R = E / I, thus (2.2 – 1,8)V / 0.36A

= 0.4 / 0.36 and it equals to **1.111 ohms**

Here the internal resistance is 1.11 ohms.

There are alternative ways to solve this question, such as:

When the cell is connected to **5 ohms**, the current flowing through the circuit is I = 2.2/(5+r) A. Where r is the cell’s internal resistance. The drop-in voltage across a resistance of **5 ohms** is

5×2.2/(5+r)=2.2–1.8 and

11=2+0.4r,

so r=9/.4 ohm.

**The third and the most accurate way of solving this is, **

- The voltage drop across the internal resistance is equal to 2.2 – 1.8 = 0.4 V.

**Current through the 5 ohms resistance=1.85=0.36A**

When two resistances are connected in series, the same current will flow through them.

**IR=0.40.36=1.11Ω**

I think now you know, how to calculate batteries’ internal resistance.

## Consider two light bulbs, one rated at 50 W and the other at 75 W, both rated at 120 V. Which bulb is the most resistant? Which bulb has the highest current?

The current must be greater to operate at higher power at the same voltage. Because current is inversely proportional to resistance, a light bulb with a higher wattage has a lower resistance.

By looking at the equation that connects power current and resistance, one can arrive at the same conclusion:

P=U2/R

When measuring the resistance of an incandescent lightbulb, one must be cautious: it will change significantly when the filament is cold compared to when it is hot. When an incandescent lightbulb is cold, it almost completely shorts out compared to when it is hot.

The lower the resistance, the higher the power consumption (for an equal Voltage). Because of the lower resistance, more current can flow for the same electrical pressure (Voltage)

**Making use of the formula Power = V2 / R**

For the 50W bulb, R=V2/P = 1202/50 = 288 Ohms.

I=P/V = 50/120 = 0.417 Amps is consumed by a 50watt bulb.

For the 75w bulb, R=V2/P = 1202 / 75 = 192 ohms.

I=P/V = 75/120 = 0.625 Amps is consumed by a 75-watt bulb.

The resistance of the 50w bulb is the highest.

**The most current is carried by the 75w bulb.**

## A 12 volt battery was connected to a 10 ohm load. The drawn current was 1.18 amps. What was the battery’s internal resistance?

To begin, you must assume that the battery’s voltage or EMF is exactly 12V. You can now solve for the internal resistance using Ohm’s Law.

Rtotal = 12 V / 1.18 A = 10.17 ohms Rtotal = V/I = 12 V / 1.18 A = 10.17 ohms

Total – Rload = 10.17 ohms – 10 ohms = 0.017 ohms

The power dissipated by a known resistance load connected across a known potential difference can be calculated by… For one minute, a 10V battery provides a resistive load of 10 ohms. What exactly is it? The 24 volt battery has an internal resistance of 1 ohm in the circuit shown, and the ammeter indicates a current of 12 A.

Or, you can do it this way

The answer to this question can be found directly in Ohm’s Law.

According to Ohm’s Law, the voltage, resistance, and current in a series-connected circuit can be calculated.

V=I⋅R

where V denotes voltage, I denotes current, and R denotes resistance

We also know that we can calculate the total resistance in a series-connected circuit by simply adding up all of the Ohms we find along the way. In this case, we have the external resistance (labeled R) and the internal resistance of the battery (which we will label r).

Because we now know the voltage (12V), current (1.18A), and external resistance (10), we can solve the following equation:

I⋅(R+r)=V

R+r=VI

r=VI−R

Substituting real numbers for our variables:

**r=121.18−10≈0.1695Ω**

## A battery’s terminal potential difference is 12 volts when connected to an external resistance of 20 ohms and 13.5 volts when connected to an external resistance of 45 ohms. What are the battery’s emf and internal resistance?

Let E be the EMF of the battery and R be the internal resistance of the battery, then for 20 ohms the current is 12/20= 0.6A and for 45 ohms the current is 13.5/45= 0.3A, so the first condition 0.6R+12=E and the second condition 0.3R+13.5=E, so solving R= 5 ohms and E= 15v.

E=15 V

r=5 Ohm

Here’s how you might go about it:

Determine the current for each circuit,

I1=0.6[A] and I2=0 .3[A]

Write an equation for each circuit using the equation U=E-I*r. There will be two equations and two variables.

Calculate E.

To find r, plug the solved value for E back into either equation.

## When the current is 1.5A, the PD of a battery is 10V, and when the current is 2.5A, the PD is 8V. What is the battery’s internal resistance?

According to the problem statement,

Vbat – Ix Ri = Pd

and it is assumed that

10 = Vbat – 1.5*Ri (Equation 1)

and

8 = Vbat – 2.5*Ri (Equation 2)

We have two linear first-order algebraic equations with two unknown quantities, which we can solve quite easily by substitution. Equation 1 is rearranged to give

Vbat = 10 multiplied by 1.5*Ri

and plugging it into Equation 2 yields

8 = (10 + 1.5*Ri) minus 2.5*Ri

Therefore

8 + (1.5–2.5) = 10

So, to determine Ri,

-2 equals -Ri

**resulting in Ri = 2 ohms**

## What is the difference between watts and volts?

**A volt is a potential energy unit**. It indicates how much energy a unit of current can provide **while an ampere is a unit to measure current. It tells us about the number of electrons flowing per second.**

A watt is a power unit that tells you how much energy is used per unit time. One watt is the amount of power provided by a one-volt supply when one amp of current flows: 1 V 1 A equals 1 W

To calculate the amount of energy used, multiply watts by time. The kilowatt-hour (kWh) is a standard unit of energy that is 1000 times the amount of energy consumed when one watt of power is used for one hour.

I think you are quite familiar with watt and Volts, and their differences.

Here is a table, showing Standard Electrical Units of measurements along with their symbols

Electrical Parameter | SI unit of measurement | Symbol | Description |

Voltage | Volt | V or E | Unit to measure Electrical PotentialV=I x R |

Current | Ampere | I or i | Unit to measure Electrical CurrentI = V/R |

Resistance | Ohms | R, Ω | Unit of DC resistanceR=V/I |

Power | Watts | W | The Unit of Power measurementP = V × I |

Conductance | Siemen | G or ℧ | The inverse of resistanceG= 1/R |

Charge | Coulomb | Q | Unit to measure electrical ChargeQ=C x V |

## Final Thoughts

Internal resistance is the resistance to the flow of current that is provided through cells and batteries. This resistance results in the generation of heat as well. Various parameters of electric current help us to find other unknown parameters.

Different practice problems lead us to a better understanding of these parameters. Different problems have been addressed previously that have helped us to find electromotive forces (emf), internal resistance, and current too.

Physics is not just understanding; it is a science of the physical parameters of our daily life. It includes current, conductance, and various laws of physics as well.

All you need to know is to practice these problems and memorize the formulas to get through your exams and any numerical problems you encounter in your life.