OHM'S LAW......

The Ohm's Law states that for a metallic conductor, at constant temperature, the ratio of the potential difference ratio V to the electrical current I is constant.This is the Ohm’s Law that can be expressed by

V/I = R or V = RI

Resistors can be combined in two kinds of arrangements: series and parallel. In the series combination the same current "I" flows along each resistor. The potential drop across each resistor, according to Ohm’s law, is V₁ = R₁I, V₂ = R₂I,….., V_n = R_nI. The overall potential difference is

V = V₁ + V₂ + ……. + V_n = (R₁ + R₂ + ….. + R_n)I

 

The system can be replaced for a single resistor R satisfying V = IR. Therefore
,

R = R₁ + R₂ + …..+ R_n

In the parallel combination the resistors are connected in such a way that the potential difference V is the same for all of them. The current through each resistor, according to Ohm’s law, is 

I₁ = V/R₁, I₂ = V/R₂, ……, I_n = V/R_n.

The total current "I" supplied to the system is

I = I₁ + I₂ +……+ I_n = (1/R₁ + 1/R₂ + …… + 1/R_n)V

The system can be replaced for a single resistor R satisfying
I = V/R. Therefore,

1/R = (1/R₁ + 1/R₂ + …… + 1/R_n) 

Question
An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-free. What is the resistance of the lamp?


The gain of potential energy occurs as a charge passes through the battery, that is, it gains a potential of V =6.0V. No energy is lost to the wires, since they are assumed to be resistance-free. By conservation of energy, the potential that was gained  must be lost in the resistor. So, by Ohm's Law:

V = I RR=V/I R = 3.0