Name: Meter Bridge
Uploaded: 2019-07-19
Duration: 7 min 2 s
Description: Measurement, by comparison, is always accurate. The video talks about a metre bridge device that helps us find the unknown resistance by comparing it with a...

Question

In figure, battery $1$ has $ε_{1} = 12.0 V$ and internal resistance $r_{1} = 0.025 Ω$ and battery $2$ has emf $ε_{2} = 12.0 V$ and internal resistance $r_{2} = 0.012 Ω$ . The batteries are connected in series with an external resistance $R$ . (a) What $R$ value makes the terminal-to-terminal potential difference of one of the batteries zero? (b) Which battery is that?

Accepted Answer

A zero terminal-to-terminal potential difference implies that the emf of the battery is equal to the voltage drop across its internal resistance, that is, $ε = i r$ .

To be as general as possible, we refer to the individual emf's as $ε_{1}$ and $ε_{2}$ and wait until the later steps to equate them $(ε_{1} = ε_{2} = ε)$ . The batteries are placed in series in such a way that their voltages add; that is, they do not oppose each other.

The total resistance in the circuit is therefore $R_{total} = R + r_{1} + r_{2}$ (where the problem tells us $r_{1} > r_{2}$ ), and the net emf in the circuit is $ε_{1} + ε_{2}$ , Since battery $1$ has the higher internal resistance, it is the one capable of having a zero terminal voltage, as the computation in part (a) shows.
(a) The current in the circuit is
$i = \frac{ε_{1} + ε_{2}}{r_{1} + r_{2} + R},$
and the requirement of zero terminal voltage leads to $ε_{1} = i r_{1},$ or
$R = \frac{ε_{2} r_{1} - ε_{1} r_{2}}{ε_{1}} = \frac{(12.0 V) (0.025 Ω) - (12.0 V) (0.012 Ω)}{12.0 V} = 0.013 Ω$

Note that $R = r_{1} - r_{2}$ when we set $ε_{1} = ε_{2}$

(b) As mentioned above, this occurs in battery $1$ .
If we assume the potential difference across battery $2$ to be zero and repeat the calculation above, we would find $R = r_{2} - r_{1} < 0,$ which is physically impossible. Thus, only the potential difference across the battery with the larger internal resistance can be made zero with suitable choice of $R$ .

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