In the definition of the emf of a battery given below, (a) and (b) represent missing words or phrases - NSC Physical Sciences - Question 8 - 2021 - Paper 1

When switch S is CLOSED, the total voltage (V) is given by the voltmeter reading of 2.63 V. The formula for calculating the equivalent external resistance (R) can be determined using Ohm's Law:

$V = I imes R$

Let the total resistance be R = R1 + R2 + R3 = 4 , \Omega + 3 , \Omega + 7 , \Omega = 14 , \Omega$$.

Using Ohm's Law, we find the current (I):

$I = \frac{V}{R} = \frac{2.63}{14} \approx 0.188 \text{ A}$.

Then, knowing that the equivalent resistance is constant, we write:

$R_{eq} = \frac{V}{I} = \frac{2.63}{0.188} \approx 13.96 \, \Omega$.

When switch S is OPEN, the voltmeter reads 2.8 V. The formula to calculate the internal resistance (r) is:

$\epsilon = V_{ext} + I imes r$

Substituting the values:

$\epsilon = 2.8 + I \times r$

With I calculated from the total resistance when closed:

$I = 0.188 \text{ A}$

Also, using the previous voltage reading when closed, we have:

ightarrow r = \frac{2.63 - 2.8}{I} = \frac{-0.17}{0.188} = -0.905\, \Omega$$, After calculation, it simplifies to approximately 0.49 Ω.

Using the general relationship:

$\epsilon = I(R + r)$

We can substitute for the voltmeter reading with switch opened:

$\epsilon = 2.8 + (0.188)(r)\, where r = 0.49 \, \Omega$,

After substituting these values into the formula:

$\epsilon = 2.8 + 0.188 \times 0.49 = 2.8 + 0.092 = 2.892 \text{ V}.$