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Question help

Hey there,

Need some help with this question.. I don't really understand the answer tbh:

A second identical battery is connected in parallel with the first one.
Describe and explain qualitatively what would happen to the voltmeter
reading if the switch remains closed. (3)

The picture showing the circuit is attached

The answers are: Voltmeter reading increases since EMF is unchanged / total resistance is reduced etc. Explain please!

Thanks in advance.

Reply 1

Please post the entire question, because I think there's something you're not telling us!

I'm guessing it's about internal resistance though. When current is flowing, the internal resistance, rr, of the battery becomes important.

Know that Vmeter=ϵIrV_{meter} = \epsilon - Ir.

Thus, the voltmeter will first read V1=I(Rr)V_1 = I(R-r). If you put another identical battery in parallel with the first, then the EMF will stay the same (6V), but because there are now two paths for the current to take, the internal resistances add like parallel resistors, giving a new effective internal resistance, rr':

1r=1r+1r[br]r=r2\\\frac{1}{r'} = \frac{1}{r} + \frac{1}{r}[br]\\r' = \frac{r}{2}

So V2=I(Rr2)V_2 = I'(R-\frac{r}{2}).

If we assume III' \approx I \dagger:

ΔV=V2V1=I(rr2)=Ir2>0[br]ΔV>0\\\Delta V = V_2 - V_1 = I(r - \frac{r}{2}) = I\frac{r}{2} > 0[br]\\\Rightarrow \Delta V > 0
Hence, the voltage increases.



\\\displaystyle \daggerThis is not a bad approximation when rR\displaystyle r \ll R:

I=ϵR+r2I' = \frac{\epsilon}{R + \frac{r}{2}} and I=ϵR+rI = \frac{\epsilon}{R + r}
Unparseable latex formula:

\\I' - I = \frac{\epsilon}{R + \frac{r}{2}} - \frac{\epsilon}{R + r}[br]\\= \frac{\epsilon}{2}(\frac{r}{R^2 + \frac{3}{2}Rr + \frac{r^2}{4}}})[br]\\\approx \frac{\epsilon}{2}(\frac{1}{\frac{R^2}{r}})[br]\\= \frac{\epsilon r}{2R^2}



Since rR2r \ll R^2, the change in current is negligible.

Reply 2

Okay, i'll post the rest.

3. A battery of e.m.f. 6.0 V is connected to a 10 Ω resistor as
shown in the circuit diagram.

(a) Define the e.m.f. of the battery.(2)

(b) When the switch is open the voltmeter reading is 6.0 V.

The internal resistance of the battery is 0.40 Ω.
Calculate the reading on the voltmeter when the switch is closed. (3)

(c) A second identical battery is connected in parallel with the first one.
Describe and explain qualitatively what would happen to the voltmeter
reading if the switch remains closed. (3)

Reply 3

Creative
Okay, i'll post the rest.

(c) A second identical battery is connected in parallel with the first one.
Describe and explain qualitatively what would happen to the voltmeter
reading if the switch remains closed. (3)

Hint. The second battery acts as a second resistor in parallel to the first...

Reply 4

okay.. so

1/Rt = 1/internal res 1 + 1/internal res2

ahh.. so the Rt is less, therefore the voltmeter reading will increase since the current will be higher V = IR.. etc

Ok i get it! Thanks everybody.