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In simple circuit is current in wire different from current in resistor as wire also opposes the flow of electrons so wire should also be added in series connection of resistor

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    $\begingroup$ I'm not sure I understand your question if there is any. What does it mean to the current here and there to be different. You can say that the amperage is different but I'm not sure this is what you are asking. $\endgroup$ – Gonenc Jun 8 '15 at 19:13
  • $\begingroup$ See if we think of a series circuit cant we say wire as also a resisitor as wires also oppose the flow of electrons $\endgroup$ – Pratyush Rohilla Jun 8 '15 at 19:17
  • $\begingroup$ Your currently accepted answer is wrong. The current in the wire and in the resistor are identical. They have incorrectly reinterpreted your question. You asked if current in the wire was different from the current in the resistor. They are identical. | Adding a wire with resistance of its own increases the total resistance so reduces the the current flow but the same current flows in the resistor and in the wire. $\endgroup$ – Russell McMahon Jun 9 '15 at 4:01
  • $\begingroup$ @PratyushRohilla You MUST stop trying to force incorrect concepts onto the circuit. It is good to ask questions but several people have both told you the answer and explained it. You are not listening to what is being said. In a series circuit ALL the current that leaves an energy source (say a battery ) flows through a wire to the load. ALL the current flows througfh the load. ALL the current flows back via another wire. ALL = ALL. $\endgroup$ – Russell McMahon Jun 9 '15 at 17:00
  • $\begingroup$ Ok russell i am ok with series but what about parallel connection. How does the current gets change in different resistors? $\endgroup$ – Pratyush Rohilla Jun 9 '15 at 17:17
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After this comment

"See if we think of a series circuit cant we say wire as also a resisitor as wires also oppose the flow of electrons"

I think your question is this:

If I say that the system has a resistor of $5\Omega$ but there is also the resistance of wire so the total resistance is greater than $5\Omega$?

Then the question is of course yes. The total resistance is greater than $5\Omega$ precisely because of the resistance of the wire. However in physics we think of an ideal world where the wires have no resistance ie made out of super conducting material. If it really troubles you that we are idealising the world too much then think when you draw a circuit that the resistance you draw as $5\Omega$ is really the sum of the resistance you drew and the resistance of the wire.

If on the other hand you are asking whether the same current flows through the resistance and the wire then the answer is a clear yes as the other answerers suggested.

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  • $\begingroup$ Thank you very much can u also tell me why total resistance is lower in parallel connection $\endgroup$ – Pratyush Rohilla Jun 8 '15 at 19:43
  • $\begingroup$ You have incorrectly reinterpreted the question. They asked if current in the wire was different from the current in the resistor. They are identical, but your answer may lead them to believe that they are different. You should clarify your answer so that it is not misleading. $\endgroup$ – Russell McMahon Jun 9 '15 at 3:59
  • $\begingroup$ I have downvoted this answer. I VERY SELDOM do this but this answer will mislead a very uncertain question-asker. Please fix it and I will upvote it. $\endgroup$ – Russell McMahon Jun 9 '15 at 4:39
  • $\begingroup$ Notice however what OP said in the comment. That's why I answered it this way. I think it's best to wait a clarification from OP about his particular question. $\endgroup$ – Gonenc Jun 9 '15 at 8:29
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    $\begingroup$ @PratyushRohilla If you had read my answer - even a small part of it , you would not need to ask that question. Read my "htdraulic pipe & fluid" analogy. ALL current flows in ALL resistors and ALL wires in a real-world series circuit. The current in ALL components and wires is IDENTICAL. ALL the same. NO difference. ALL = ALL. SAME = SAME . $\endgroup$ – Russell McMahon Jun 9 '15 at 16:53
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The current will be the same in the wire and the resistor if they are in series. This is ensured by conservation of charge - you cannot have more electrons leaving a component than entering it. (Kirchoff's current law)

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  • $\begingroup$ So you are basically saying that the current in the wire and the resisitor are same $\endgroup$ – Pratyush Rohilla Jun 8 '15 at 19:13
  • $\begingroup$ @PratyushRohilla YES - the same current flows in both. $\endgroup$ – Russell McMahon Jun 9 '15 at 4:03
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In simple circuit is current in wire different from current in resistor as wire also opposes the flow of electrons so wire should also be added in series connection of resistor

IF you follow this answer step by step the situation should be MUCH clearer to you.
IF you just skim this answer it will completely confuse you.

Short summary:

Read my "hydraulic pipe & fluid" analogy.
Just as in a hose & fluid series connected system ALL the fluid flows through ALL the components, so too in an electric circuit

  • ALL current flows in ALL resistors and ALL wires
    in a real-world series circuit.

  • The current in ALL components and wires is IDENTICAL.
    ALL the same.
    NO difference.

  • ALL = ALL.
    SAME = SAME .

Note to editors: The language used is intended to be simple and clear. Some statements are very "redundant". This is intentional. Editing to remove certain errors is OK. Editing which destroys the simple and structured approach will be 'rolled back'.


A series circuit of wire plus load is similar to a pipe feeding hydraulic oil to a hydraulic motor (eg a power steering motor in a car's power steering unit which is fed by hydraulic fluid from the steering pump).
Imagine a hydraulic pump with outlet (high pressure), inlet (low pressure), a hydraulic steering motor, a feed pipe from pump outlet to the motor and a return pipe from motor to pump inlet.

ALL the fluid which leaves the power steering pump outlet flows in the feed pipe
All the fluid which flows in the feed pipe flows into the motor.
All the fluid which flows out of the motor flows into the return pipe.
All the fluid which flows in the return pipe flows back into the pump inlet.

ALL the fluid flows = current are the same.
The pressure drops (voltages) across the pipes take pressure away from the motor and so we wish to keep pipe pressure drops low.
Total pump outlet pressure = feed-pipe-pressure_drop + pump_pressure_drop + return_pipe_pressure_drop.

Effect of wire on total resistance

The resistance in a "series" circuit is the sum of all the individual resistances.
So if a wire has resistance and a load has resistance the total resistance is the sum of the two
$$R_{total} = R_{load} + R_{wire}$$

Effect of wire on current flowing

Current flowing through a series circuit is the same throughout. Whatever flows through a connecting wire will also flow in the load.

However, for a fixed input voltage, increasing the total resistance will reduce current flow.
Current flow can be calculated by a form of Ohms law
Current = Voltage/Resistance
$$I = V / R_{total} $$ Here $R_{total} = R_{load} + R_{wire}.$
For a give $V$, if $R_{total}$ increases then $I$ will be reduced.
So if $R_{wire}$ increases $I$ will be decreased.
But the current $I$ in wire and load will be the same as each other.

Effect of wire on voltage drops

Rearranging Ohm's law we see that for a given current flow, the voltage drop is proportional to resistance
Voltage drop = Current x Resistance
$$V = I \times R $$ Here $$V_{total} = I \times R_{total}$$
$$V_{total} = I \times (R_{load} + R_{wire}) $$ $$V_{total} = (I \times R_{load}) + (I \times R_{wire})$$

That is, the SAME current flows through both wire and load and the voltage drop in each is proportional to their resistances. If we have fixed voltage $V$ available as input, increasing $R_{wire}$ will increase $V_{wire}$, reduce $I_{total}$ and reduce $V_{load}$.

Usually wiring is arranged so $R_{wire}$ is much less than $R_{load}$ so that $V_{wire}$ is much less than $V_{load}$.


Simple example

If $$V = 30 V $$ $$R_{load} = 10 \Omega$$ $$R_{total} = R_{load} + R_{wire}$$
short version: $$R_t = R_l + R_w$$

$$I = V/R $$ $$R = V/I $$ $$V = IR$$

(a) If $R_w = 0 $ $$R_{total} = R_l + R_w = 10\Omega$$ $$I = V/R = V/R_t = V/10 = 3A$$ $$V_{load} = V_{total} = 30V $$ or $$V_{load} = IR = I \times R_{load} = 3A \times 10\Omega = 30V$$

(b) If $R_{wire} = 5 \Omega$
$$R_t = R_l+R_w = 10\Omega + 5\Omega = 15 \Omega $$ $$I = V/R = V/R_t = 30V/15\Omega = 2 A$$ Current is now $2A$ compared to $3A$ in (a) above.
$$V_{load} = IR = I \times R_{load} = 2A \times 10\Omega = 20V $$ $$V_{wire} = IR = I \times R_{wire} = 2A \times 5\Omega = 10 V$$ $1/3$ of voltage is dropped by wire
$1/3$ of $R_{total}$ is $R_{wire}$

(c) If $R_{wire} = 20 \Omega$ (very bad situation) $$R_t = R_l+R_w = 10\Omega + 20\Omega = 30\Omega $$ $$I = V/R = V/R_t = 30V/30\Omega = 1 A$$ Current is now $1A$ compared to $3A$ in (a) above.
$$V_{load} = IR = I \times R_{load} = 1A \times 10\Omega = 10V$$
$$V_{wire} = IR = I \times R_{wire} = 2A \times 20\Omega = 20 V$$ $2/3$ of voltage is dropped by wire
$2/3$ of $R_{total}$ is $R_{wire}$

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