Question

How much energy is given to each coulomb of charge passing through a 6 V battery?

Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.​

Answer 1

Q1

We know that the potential difference between two points is given by the equation,

V = W/Q, where,

W is the work done in moving the charge from one point to another

Q is the charge

From the above equation, we can find the energy given to each coulomb as follows:

W = V × Q

Substituting the values in the equation, we get

W = 6V × 1C = 6 J

Q2

Resistance is given by the equation,

R = ρ l/A

where,

ρ is the resistivity of the material of the wire,

l is the length of the wire

A is the area of the cross-section of the wire.

From the equation, it is evident that the area of the cross-section of wire is inversely proportional to the resistance. Therefore, thinner the wire, more the resistance and vice versa. Hence, current flows more easily through a thick wire than a thin wire.

Q3

The amount of heat generated can be calculated using the Joule’s law of heating, which is given by the equation

H = VIt

Substituting the values in the above equation, we get,

H = 100 × 5 × 30 = 1.5 × 104 J

The amount of heat developed by the electric iron in 30 s is 1.5 × 104 J.

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V of battery.