Question

how should be the resistors connected in ohm's law???

Answer 1

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated by using the equation V=IR, where I is current in amps (A) and R is the resistance in ohms (Ω).

So the voltage drop across R1 is V1=IR1, across R2 is V2=IR2, and across R3 is V3=IR3. The sum of the voltages would equal: V=V1+V2+V3, based on the conservation of energy and charge. If we substitute the values for individual voltages, we get:

V=IR1+IR2+IR3V=IR1+IR2+IR3

or

V=I(R1+R2+R3)V=I(R1+R2+R3)

This implies that the total resistance in a series is equal to the sum of the individual resistances. Therefore, for every circuit with N number of resistors connected in series:

RN(series)=R1+R2+R3+…+RN.RN(series)=R1+R2+R3+…+RN.

Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up.

Since voltage and resistance have an inverse relationship, individual resistors in series do not get the total source voltage, but divide it. This is indicated in an example of when two light bulbs are connected together in a series circuit with a battery. In a simple circuit consisting of one 1.5V battery and one light bulb, the light bulb would have a voltage drop of 1.5V across it. If two lightbulbs were connected in series with the same battery, however, they would each have 1.5V/2, or 0.75V drop across them. This would be evident in the brightness of the lights: each of the two light bulbs connected in series would be half as dim as the single light bulb. Therefore, resistors connected in series use up the same amount of energy as a single resistor, but that energy is divided up between the resistors depending on their resistances.