Physics Internal Resistance In Series

I'm not sure how to go about the question:
Two cells are connected in series, Each has an electric motor force of 1.4V and internal resistance of 0.38 ohms. They are connected across an electric circuit of resistance 1.8 ohms:
Calculate the voltage across the circuit
Calculate the voltage across the terminals of each cell

I tried them both but the method seemed a bit odd:
I totalled the input voltage to 2.8V and the total resistance to 0.76 + 1.8 = 2.56 ohms
I tried doing the potential divider formula (R1/Total R) x 2.8 which gave me 1.96 but my teacher said this was wrong?

For the latter I apparently got it wrong as well by doing a similar method by treating the voltage across across each cell as a potential divider and using the formula again getting 0.38 / (1.8 + 0.76) * 2.8 which gave me roughly 0.42V

Where did I go wrong?

Reply 1

Shaanv

Just to confirm the total resistance of the whole circuit is 2.56 ohms including the resistance of the cells

Reply 2

AmmarTa

Original post by Shaanv

Just to confirm the total resistance of the whole circuit is 2.56 ohms including the resistance of the cells

Yes

Reply 3

Shaanv

Original post by AmmarTa

Yes

Just reading ur method confuses me slightly.

The way i would approach the question is i think simpler.

Find the current, as it is the same all around the series circuit. To do this do emf/resistance.

For the voltage across each cell use V=IR with the resistance of one cell.

As u have two cells double this and take it away from the emf and u are left with the circuit voltage.

U can also find the voltage of the circuit first in a similar way.

For the circuit i get 1.97 V (2dp)
For each cell i get 0.42 V (2dp)

So our answers are very similar maybe they disagree with ur method as it seems more complicated than it needs to be.

Could u post a pic of the actual question.

Reply 4

AmmarTa

Original post by Shaanv

Sure, also I got the same answers but I rounded a bit dodgy on the first one :P
One moment

Reply 5

AmmarTa

Original post by Shaanv

Question 7

(edited 6 years ago)

15092005635341330750064.jpg539.7KB

Reply 6

Shaanv

Original post by AmmarTa

Question 6

The only other thing i could think of is that maybe the circuit resistance includes the resistance of each cell as well.

Reply 7

AmmarTa

Original post by Shaanv

The only other thing i could think of is that maybe the circuit resistance includes the resistance of each cell as well.

It was Q7 sorry

Reply 8

Shaanv

Original post by AmmarTa

It was Q7 sorry

I figured and what i said applies to q7

Reply 9

AmmarTa

Original post by Shaanv

I figured and what i said applies to q7

I suppose it should be right then tbh, my teacher tends to get things wrong

Reply 10

Shaanv

Original post by AmmarTa

I suppose it should be right then tbh, my teacher tends to get things wrong

I believe so

Reply 11

AmmarTa

Original post by Shaanv

I believe so

If it's not too much trouble can you explain Q6 too please

Reply 12

Eimmanuel

Study Forum Helper

Original post by AmmarTa

I don't think your method is wrong. I believe is the significant figures that get you wrong. All the values are only quoted to 2 s.f., but you state your answer as 3 s.f.

Original post by AmmarTa

For the latter I apparently got it wrong as well by doing a similar method by treating the voltage across across each cell as a potential divider and using the formula again getting 0.38 / (1.8 + 0.76) * 2.8 which gave me roughly 0.42V

Where did I go wrong?

This part is definitely wrong. The voltage across the external circuit is provided by the two cells, so each cell should have (1.968 V)/2 = 0.98 V.

Or you can find the current through each cell and use the following equation to find the voltage across each cell.

V = E - Ir.

E is the emf which is 1.4 V.
r is the internal resistance which is 0.38 ohms.
V is the voltage across each cell.

Reply 13

AmmarTa

Original post by Eimmanuel

That makes sense, I forgot the equations

Reply 14

Shaanv

Original post by AmmarTa

If it's not too much trouble can you explain Q6 too please

For part a if u know that 0.81V is supplied by the cell and its emf is 1.4V then the difference between these two must be the lost volts.

For part b first find the current. I did this by considering the wire. As it has a resistance of 7.3 ohms and the voltage across the wire must be 0.81V u can work out current using V=IR.
Then we can use V=IR again but considering the source this time, to work out interntal resistance.

Part c: the power dissipated by the wire is P=I^2*R and the power supplied by the cell is P=IV where V is the emf.

Reply 15

Dat1Guy

Your method for question 7 was right, although you could have just found the current throughout the circuit and looked at each resistor independently to find the voltage. But the potential divider equation does work, you just misused the calculator.