**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 V**

_{1}= R_{1}I, V_{2}= R_{2}I,….., V_{n}= R_{n}I. The overall potential difference is**V = V**

_{1}+ V_{2}+ ……. + V_{n}= (R_{1}+ R_{2}+ ….. + R_{n})I

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

**,**

**R = R**_{1}+ R_{2}+ …..+ 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**

**I**

_{1}= V/R_{1}, I_{2}= V/R_{2}, ……, I_{n}= V/R_{n}.**The total current "I" supplied to the system is**

**I = I**

_{1}+ I_{2}+……+ I_{n}= (1/R_{1}+ 1/R_{2}+ …… + 1/R_{n})V**The system can be replaced for a single resistor R satisfying**

I = V/R. Therefore,

I = V/R. Therefore,

__1/R = (1/R___{1}+ 1/R_{2}+ …… + 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 R**

**R=V/I**

**R = 3.0**

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