EMFWatch

#1
Hey Guys,
I have been trying out this question to test my knowledge and understanding of the topic but i don't seem to be able to solve it.

Question:
A cell is connected in series with an 8.0ohms resistor and a switch. A high resistance voltmeter is connected across the cell and reads 3.6V when the switch is open and 3.2V when the switch is closed . Calculate (a) EMF (b) the internal resistance

My attempt:
So i drew out the diagram and got to the conclusion that the voltage of the resistor is 3.2V because thats the voltage going through when the switch is closed. Hence i worked out the current as 0.4A. Then i worked out the equation `E-0.4r=3.6' . Im pretty sure that was right so i formed another equation `E/(8+r) = 0.4 ' But then when i rearrange it and try to eliminate one factor i got `0=0.4'

0
4 years ago
#2
(Original post by marcoo2857)
Hey Guys,
I have been trying out this question to test my knowledge and understanding of the topic but i don't seem to be able to solve it.

Question:
A cell is connected in series with an 8.0ohms resistor and a switch. A high resistance voltmeter is connected across the cell and reads 3.6V when the switch is open and 3.2V when the switch is closed . Calculate (a) EMF (b) the internal resistance

My attempt:
So i drew out the diagram and got to the conclusion that the voltage of the resistor is 3.2V because thats the voltage going through when the switch is closed. Hence i worked out the current as 0.4A. Then i worked out the equation `E-0.4r=3.6' . Im pretty sure that was right so i formed another equation `E/(8+r) = 0.4 ' But then when i rearrange it and try to eliminate one factor i got `0=0.4'

Hello and welcome to TSR.

The equation should be E - 0.4r = 3.2

The open circuit battery voltage is 3.6V (i.e. this is the voltage developed across an infinite series load resistance)

With the switch closed, the voltage developed across the 8 ohms resistor is 3.2V

Therefore the current I = V/R = 3.2/8 = 0.4A

According to Kirchoff: 3.6V - 3.2V = 0.4V = potential developed across the battery internal resistance.

Since the current must be the same (series path) then the internal resistance is

r = Vr/I = 0.4V/0.4A

r = 1 ohm.
0
#3
(Original post by uberteknik)
The open circuit battery voltage is 3.6V (i.e. this is the voltage developed across an infinite series load resistance)

With the switch closed, the voltage developed across the 8 ohms resistor is 3.2V

Therefore the current I = V/R = 3.2/8 = 0.4A

According to Kirchoff: 3.6V - 3.2V = 0.4V = potential developed across the battery internal resistance.

Since the current must be the same (series path) then the internal resistance is

r = Vr/I = 0.4V/0.4A

r = 1 ohm.
Thanks! but can you please explain the kerchief law that you used?
0
4 years ago
#4
(Original post by marcoo2857)
Thanks! but can you please explain the kerchief law that you used?
You need to learn both Kirchoff's voltage and current rules (laws).

KVL = Kirchoff's Voltage Law = The sum of the voltage drops around a closed circuit is equal to the supply voltage.

You made an error with the equation E - 0.4r = 3.6

Using KVL it should be E - 0.4r = 3.2

i.e. 3.6 - 0.4r = 3.2

0.4r = 0.4

r = 0.4./0.4
0
#5
(Original post by uberteknik)
You need to learn both Kirchoff's voltage and current rules (laws).

KVL = Kirchoff's Voltage Law = The sum of the voltage drops around a closed circuit is equal to the supply voltage.

You made an error with the equation E - 0.4r = 3.6

Using KVL it should be E - 0.4r = 3.2

i.e. 3.6 - 0.4r = 3.2

0.4r = 0.4

r = 0.4./0.4
ahhh okay but what i don't understand is that if `E' is supposed to be `EMF' then why would you put '3.6V' when thats the voltage you get after subtracting the lost voltage from the EMF?
0
4 years ago
#6
(Original post by marcoo2857)
ahhh okay but what i don't understand is that if `E' is supposed to be `EMF' then why would you put '3.6V' when thats the voltage you get after subtracting the lost voltage from the EMF?
Make sure you read the question.

3.2V is the voltage developed across the resistor after the switch is closed.

3.6V is the open circuit voltage (no load = infinite resistance, zero current). This voltage is in effect, developed across an infinite resistance which itself is in series with both the 8 ohms resistor and the internal battery resistance.

This means in the open circuit condition, NO potential is developed across the series
(r + R) combination.
0
#7
(Original post by uberteknik)
Make sure you read the question.

3.2V is the voltage developed across the resistor after the switch is closed.

3.6V is the open circuit voltage (no load = infinite resistance, zero current). This voltage is in effect, developed across an infinite resistance which itself is in series with both the 8 ohms resistor and the internal battery resistance.

This means in the open circuit condition, NO potential is developed across the series
(r + R) combination.
THANKYOU! that justifies everything you said now
0
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