# Electrical Formulas

Electricity is an integral part of our lives. In our daily life, we need electricity to run lights, fans, air conditioners, computers, etc. So, what is electricity? The flow of a charge through a conductor is referred to as electricity. The charge usually flows from a positive terminal (anode) to a negative terminal (cathode). To understand the concept of electricity better, we need to understand the various basic parameters, like the voltage, current, resistance, and conductivity, and also the relationship between them.

**Current**

Electric current is generally defined as the rate at which charged particles flow through a conductor. Current is measured in terms of amperes, where 1 ampere is defined as 1 coulomb of charge passing through the cross-section every 1 second.

Current (I) = Q/tWhere Q is a charge,

and T is the time taken.

**Voltage**

The force that causes an electric charge to move is called voltage. In simple words, voltage is defined as the potential difference per unit charge between two points. It is measured in terms of volts (V).

Voltage (V) = W/QWhere W is the work done,

and Q is the electric charge.

We can also calculate voltage by using Ohm’s law.

Voltage (V) = IRWhere I is the current flowing through the circuits,

and R is the resistance of the electrical circuit.

**Resistance**

The property of a conductor which resists the flow of electric current through it is referred to as resistance. A resistor is a component of the circuit that opposes the flow of current in a circuit. The resistance of a conductor depends on the nature of the material used to build that conductor, its length, and its area of cross-section, and it is measured in terms of ohms (Ω).

The resistance of a conductor (R) is directly proportional to its length (l) and inversely proportional to its area of cross-section (A).

So, R ∝ l and R ∝ 1/A

⇒ R ∝ l/A

⇒ R = ρ l/A

Where ρ is the resistivity of the conductor which depends on the material of the conductor.

Resistance (R) = ρl/AWhere ρ is the resistivity of the conductor,

l is the length of the conductor,

and A is the area of cross-section of the conductor

### **Power**

In an electric circuit, the rate of transfer of electrical energy per unit of time is defined as electric power. It is measured in terms of watts (W) and is mathematically represented as the product of potential difference and current.

Power (P) = VIWhere V is the voltage or potential difference,

and I is electric current.

From Ohm’s law, we have** V = IR**

So, Power (P) = IR × I = I^{2}R

I = V/R

So, P = V^{2}/R

Power (P) = I^{2}R

(or)

Power (P) = V^{2}/RWhere,

I is current,

R is resistance,

V is voltage.

**Conductivity**

The electrical conductivity of a substance is referred to as the measure of the transfer of electric current through it. For a given potential difference, the current density of a substance is high if the electrical conductivity of a substance is high. It is represented by “σ”, and is measured in terms of Siemens per meter (Sm^{-1}).

Conductivity (σ) = 1/ρWhere, ρ is the resistivity.

### Sample Problems:

**Problem 1: Calculate the amount of charge flowing through the electric lamp if the lamp draws a current of 1 A and lights for 6 hours. **

**Solution:**

Given data,

Current (I) = 2 A

Time (t) = 6 hr

t = 6 × 3600 sec = 21600 sec

We know that,

Charge (Q) = I × t

Q = 2 × 21600 = 43200 C

Hence, the amount of charge flowing through the electric lamp is 43200 C.

**Problem 2: What is the power generated by a circuit whose voltage is 15 V and resistance is 45 Ω?**

**Solution:**

Given data,

Potential difference (V) = 15 V

Resistance (R) = 45 Ω

We know that,

Power (P) = V

^{2}/RP = (15)

^{2}/45 = 225/45P = 5W

Hence, the power generated by the given circuit is 5W.

**Problem 3: Determine the magnitude of current flowing through the electric iron box if it has a potential difference of 250 V and a resistance of 80 Ω.**

**Solution:**

Given data,

Potential difference (V) = 250 V

Resistance (R) = 80 Ω

We know that,

Current (I) = V/R

I = 250/80 = 3.125 A

Hence, the current flowing through the electric iron box is 3.125 A.

**Problem 4: Find out the resistance value of a resistor that has a potential difference of 12V and radiates heat and produces thermal energy at a rate of 6W. **

**Solution:**

Given data,

Potential difference (V) = 12 V

Power (P) = 6W

We know that,

Power (P) = V

^{2}/R⇒ R = V

^{2}/P⇒ R = (12)

^{2}/6 = 144/6⇒ R = 24 Ω

Hence, the resistance value of the resistor is 24Ω.

**Problem 5: How much power does the inverter draw if the battery of an inverter functions at 24 V and draws a current of 0.4 A?**

**Solution:**

Given data,

Potential difference (V) = 24 V

Current (I) = 0.4 A

We know that,

Power (P) = VI

⇒ P = 24 × 0.4 = 9.6 W

Hence, the power drawn by the inverter is 9.6 W

**Problem 6: What is the resistivity of a conductor if its resistance is 10 Ω, its length is 3 m, and its area is 3.5 mm ^{2}?**

**Solution:**

Given data,

Resistance (R) = 10 Ω

Area of the conductor (A) = 3.5 mm

^{2}= 3.5 × 10^{-6}m^{2}Length of the conductor (l) = 3 m

We know that,

Resistance (R) = ρl/A

Resistivity (ρ) = RA/l

ρ = (10 × 3.5 × 10

^{-6})/(3)ρ = 11.67 × 10

^{-6}Ω-mHence, the resistivity is 11.67 × 10

^{-6}Ω-m.

**Problem 7: Determine the conductivity of a conductor with a resistivity of 1.8 × 10 ^{-6} Ω-m. **

**Solution:**

Given data,

Resistivity (ρ) = 1.8 × 10

^{-6}Ω-mWe know that,

Conductivity (σ) = 1/ρ

ρ = 1/(1.8 × 10

^{-6})= 0.556 × 10

^{6}= 5.56 × 10^{5}Sm^{-1}Hence, the conductivity is 5.56 × 10

^{5}Sm^{-1}.