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.
Answers
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.