Question

In Figure (a), both batteries have emf ε = 1.30 V and the external resistance R is a variable resistor. Figure (b) gives the electric potentials V between the terminals of each battery as functions of R: Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2. The horizontal scale is set by Rs = 0.300 Ω. What is the internal resistance of (a) battery 1 and (b) battery 2?

0.65 R 0 R 0.390 R (22) (6) (a)

Answer 1

The voltage across the terminals of a battery is its EMF minus the potential due to the internal resistance. Because the internal resistance reduces the voltage.

V battery = E - iRinternal

Thus we model a real battery as an ideal voltage source in series with its internal resistance. Lets call the internal resistance of EMF ε_i = r_i

Then, applying the kirchoff’s law in this loop we get

-iR+ (E2 - ir2) + (Ei - iri) = 0

The terms in parenthesis are the voltages of the batteries. (which we can also see plotted in the graph as a function of R.

-iR + V1 + V2 = 0

Lets use the point where R = ½ R_s = 0.150 Ω. I read from the graph that V₁ = 0.52V

0.65 1 R V(V) 1 3 - - 0 RE -0.390 R (12) (b) (a)

Because 5 divisions = 0.65

⇒ 1 division = 0.65/5 = 0.13

⇒ 4 divisions = 0.13 x 4 = 0.52

and V₂ = 0V. Now we can write down the loop equation using these values:

-iR + V1 + V2 = 0


$-i(0.150 \Omega ) + 0.52V+0V=0$

Therefore, current at that moment can be calculated as:

$\left | i \right |=\frac{0.52V}{0.150 \Omega} = 3.467 A$

From this we can calculate the internal resistances:

$\varepsilon _1-ir_1 = 0.52$

1.3 - iri = 0.52

iri = 1.3 -0.52 = 0.78

$r_1 =\frac{0.78}{3.467}=0.225 \Omega$

Similarly,

$\varepsilon _2-ir_2 = 0$

$r_2=\frac{1.3}{3.467}=0.375 \Omega$

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