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I would like to know why by adding two resistors in series we would get a sum of it's resistances.I know by Ohm's Law that since $$R=\frac {V}{I}$$ it's certain that you would attain the equation $$R_t=R_1+R_2$$

But I would like to have a practical definition on how resistances in series produce sum of it's resistances.

For example in the image below the point that confuses me is that how does the first resistor's resistance influences the second resistor's resistance. Because if we would try to find the current through the second resistor what I expect is that it (current) of the second resistor should be found by dividing only the resistance of the second resistor with the voltage through the circuit but unfortunately it is $$I_2=\frac{V}{R_1+R_2}$$

Could anyone help me understand why my expectation is wrong?

enter image description here

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  • \$\begingroup\$ Voltage doesn't "flow", current does. \$\endgroup\$ – pjc50 Nov 26 '14 at 13:08
  • \$\begingroup\$ @pjc50:Oh sorry.First I wrote it was current then changed to voltage and forgetted about the 'flow'. \$\endgroup\$ – justin Nov 26 '14 at 13:09
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    \$\begingroup\$ @justin . I think you are confusing ohms law with resistance. R = V / I is a mathematical formulation of Ohms law. What the law actually states is "Current flowing in a circuit is proportional to the voltage applied across its terminals" . It says nothing about resistance. Resistance enters merely as a proportionality constant, and physicists have TERMED it as resistance . And constants are , well , constant. So they do not influence each other. Their value remains constant just like ...constants. \$\endgroup\$ – Plutonium smuggler Nov 26 '14 at 14:57
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how does the first resistor's resistance influences the second resistor's resistance.

It doesn't. However, the resistance of the first resistor influences the voltage across the second resistor.

Clearly, the resistors in the diagram are series connected thus the current through each resistor is identical.

$$I_1 = I_2 = I$$

By Ohm's law, we have

$$V_1 = I_1 \cdot R_1 = I \cdot R_1$$

$$V_2 = I_2 \cdot R_2 = I \cdot R_2$$

By KVL, we have

$$6V = V_1 + V_2 = I \cdot R_1 + I \cdot R_2 = I \cdot (R_1 + R_2)$$

Thus

$$I = \frac{6V}{R_1 + R_2}$$

In other words, the current \$I\$ depends on the sum of the resistances (series connected resistances add).

The voltage across the second resistor can now be written as

$$V_2 = I \cdot R_2 = \frac{6V}{R_1 + R_2} \cdot R_2 = 6V \frac{R_2}{R_1 + R_2} $$

and so, as first stated, the resistance of the first resistor influences the voltage across the second resistor.

Similarly

$$V_1 = I \cdot R_1 = \frac{6V}{R_1 + R_2} \cdot R_1 = 6V \frac{R_1}{R_1 + R_2}$$

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  • \$\begingroup\$ Thanks for your anwser.I think it might be difficult to give a practical explanation for this question isn't it? \$\endgroup\$ – justin Nov 27 '14 at 11:17
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There is one critical error in your reasoning:

[The current] of the second resistor should be found by dividing only the resistance of the second resistor with the voltage through the circuit

The "voltage through the circuit" is not a well defined concept. Ohm's law for a resistor states that \$V = R \cdot I\$, but you need to be very careful to use the voltage and current appropriate for the specific resistor.

You can imagine that from the point of view of R2, all you know about are your own two ends ("terminals"), but you can't see anything else going on farther away in the circuit. In this case, R2 can't "see" the voltage on the positive side of the 6 V source because it is on the other side of R1.

Once you use the correct voltages in Ohm's law, the reason why the effective resistance of series resistors is a summation is derived in Alfred Centauri's answer. Note that he talks about the "voltage across the second resistor," not the "voltage through the circuit."

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  • \$\begingroup\$ Thanks for your correction.But I can't get what you meant by " R2 can't "see" the voltage on the positive side of the 6 V source because it is on the other side of R1.".I understood the literal meaning but couldn't get why you used it in this context. \$\endgroup\$ – justin Nov 27 '14 at 11:25
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You only need to know the voltage across the resistor itself. In your example V2 is the voltage across your second resistor so if you find V2 you will notice that the same current will flow if you take 6 Volts divided by 1K +220 as does if you take V2 going into 1K. for your excersize ill let you figure out what V2 is, a big hint is to use voltage division formula.

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  • \$\begingroup\$ Do you meant to say that the resistance 220 $\Ohm$ is negligible with 1K $\Ohm$? \$\endgroup\$ – justin Nov 27 '14 at 11:29
  • \$\begingroup\$ No. Im saying that there are different voltages across each resistor. the two resistors in series have the main supply. while the resistors independently have their own specific voltage drop across of them which must add to the main supply and ensure that the same current goes through both resistors. this gives you enough information to solve for the circuit with the information provided. \$\endgroup\$ – Zayzoon Nov 27 '14 at 21:49
  • \$\begingroup\$ The current through each resistor is given by the voltage on its terminals alone. Conversely the voltage on its terminals is given by the current and its resistance. Stack up all the resistor voltages to get a total voltage across the whole string, given by the sum of the individual resistor voltges. \$\endgroup\$ – Neil_UK Oct 31 '15 at 18:23

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