# Circuit with resistors in series and two power supplies

• 743344
In summary, Kirchoff's voltage law can be used to calculate the voltage drop across a resistor in a circuit. The student was incorrect in their original attempt, and correctly solved the problem after redrawing the circuit with resistors in one leg and batteries in another.
743344

## Homework Statement

There is a current of 0.25 A in the circuit of the figure

What is the value of the resistance R?

V = IR

## The Attempt at a Solution

since the circuit is in series:

V = 6+ 12 = 18V
Req = R + 6 + 12
V = IReq
Req = 18/0.25 = 72

72 = 18+ R
R = 54

but the answer is wrong !
help

current of the whole circuit =0.25A

I=V/R = (12-6)/(18+R) = 6/(18+R)
4.5+0.25R=6
0.25R=1.5
R=6 ?
I am not sure ...

I'm sure you know Kirchoff's voltage law - "The sum of all voltage drops around any closed loop must be equal to zero.

Applying Kirchoff's voltage law gives ->
0 = Power supply driving current (largest voltage) - Power supply opposing flow of current (smallest voltage) - Voltage drop across resistance in circuit (6Ohm, 12Ohm and ROhm).

0 = 12 - 6 -(6*0.25) - (12*0.25) - 0.25R
Thus..
0 = 6 - 1.5 - 3 - 0.25R
So..
0= 1.5 - 0.25R
0.25R = 1.5
R = 6Ohms

Ohm's law is something in this world i guess...

Ohm's law is something i guess.

Ohms law is something - but in this case Kirchoff's voltage law is something also.

Utility of the either one counts.

My take on this is that because there is a charge flow from the 12V battery and the potential of this battery is higher than the 6V, the 6V does not release any electrons at all and acts as a conductor.

the circuit is an illusion.

the effective voltage is not 18 volts, it is 6 volts.

redrawing circuit with resistors in one leg and batteries
in another gives a better understanding of actual circuit.

6 ohm 12 ohm Rx
+----VVVV----VVVV----VVVV---+
| |
__|__ + | E/I = Rt
___ 12 V | 6V/.25A = 24 OHM total
_____ | therefore;
___ | 6 + 12 + Rx = 24 OHM
| | Rx = 6 OHM
| 6 volts effective |
_|_ |
_____ |
___ |
_____ 6 V |
| + |
| |
+-------------------------------------+

please excuse previous post. seems that character graphics can not be used on this forum.

repeating text of previous post and adding a graphic to show redrawn circuit.

++++
the circuit is an illusion.

when 2 batteries connected with like polarities connected cause lesser voltage to be subtracted
from higher voltage. therefore, the effective voltage is not 6 volts + 12 volts = 18 volts.

it is 12 volts - 6 volts = 6 volts.

attached redrawing of circuit with resistors in one leg and batteries in another leg
gives a better of what is actual circuit.

using formula,
E/I = Rt
6V/.25A = 24 OHM total
therefore;
6 + 12 + Rx = 24 OHM
Rx = 6 OHM

geleem said:
when 2 batteries connected with like polarities connected cause lesser voltage to be subtracted
from higher voltage. therefore, the effective voltage is not 6 volts + 12 volts = 18 volts.

it is 12 volts - 6 volts = 6 volts.

You are right. A person should go around the loop in one direction, say, arbitrarily choosing clockwise. Then we would encounter one battery in the polarity plus-minus and we would encounter the other battery in the polarity minus-plus. Therefore the batteries oppose one another. We have to subtract the batteries, not add them.

mikelepore said:
A person should go around the loop in one direction, say, arbitrarily choosing clockwise.

this is true. considering as a loop is one way to look at it.

when i first looked at circuit, i saw reason "743344"'s formula did not work just from fact being that
'load' side, [normal of convention of supply to left, load to right of a circuit] had an opposing voltage.

"mysqlpress" replied with correct answer in formula without a corrected drawing.

"Mike Cookson" replied with an implication of "Kirchoff's voltage law", no drawing.

24 Dec 2009, 06:48 PM UTC, i attempted an 'ascii graphic' of circuit, which failed due to
compressing of spaces used.

after seeing this, i submitted again, 24 Dec 2009, 07:46 PM UTC, with an 'attached' drawing,
which for some reason or other, did not get shown. [i just now tried an edit and was able to get
attachment to save]

all in all, another way to spend December 24th UTC.

happy Christmas to all.

#### Attachments

• series_circuit-0021.png
15.3 KB · Views: 600
Last edited:

## What is a circuit with resistors in series and two power supplies?

A circuit with resistors in series and two power supplies is a type of electrical circuit where the resistors are connected end-to-end in a single path and there are two separate power sources providing voltage to the circuit.

## What is the purpose of having resistors in series?

The purpose of having resistors in series is to limit the current flow in a circuit, as well as to divide the voltage evenly between the resistors. This allows for more precise control of the flow of electricity and avoids overloading any one component.

## How do you calculate the total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of all the individual resistors. This can be calculated using the formula R = R1 + R2 + R3 + ... + Rn, where R is the total resistance and R1, R2, R3, etc. are the individual resistances.

## What happens to the current in a series circuit if one resistor is removed?

If one resistor is removed from a series circuit, the total resistance in the circuit decreases. As a result, the current in the circuit will increase because there is less resistance to impede the flow of electricity.

## How do the power supplies in a circuit with resistors in series affect the overall voltage?

In a circuit with resistors in series and two power supplies, the total voltage is divided between the power supplies and the resistors. This means that the voltage across each resistor will be less than the total voltage provided by the power supplies. Additionally, the voltage from the power supplies will add together to equal the total voltage in the circuit.

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