The potential difference across the terminals of a battery is 8.70 VV when there is a current of 1.47 AA in the battery from the negative to the positive terminal. When the current is 3.42 AA in the reverse direction, the potential difference becomes 10.50 VV.(a) What is the internal resistance of the battery? (b) What is the emf of the battery?

To solve for the internal resistance of the battery \( r \) and the electromotive force (emf) \( \mathcal{E} \), we'll first use the relationship given by:

\[ V = \mathcal{E} - Ir \]

when the current flows from negative to positive (conventional current flow), and:

\[ V = \mathcal{E} + Ir \]

when the current flows from positive to negative (the reversed direction).

Here, \( V \) is the terminal voltage, \( I \) is the current, \( \mathcal{E} \) is the emf, and \( r \) is the internal resistance.

Let's plug in the values for each case.

(a) For the first case with 1.47 A current and 8.70 V potential difference, we have:

\[ 8.70 = \mathcal{E} - (1.47)r \] \[ 1.47r = \mathcal{E} - 8.70 \] ---(1)

(b) For the second case with 3.42 A current (reversed) and 10.50 V potential difference, we have:

\[ 10.50 = \mathcal{E} + (3.42)r \] \[ 3.42r = 10.50 - \mathcal{E} \] ---(2)

Now, we have two equations with two unknowns. By solving these simultaneous equations, we can find both \( \mathcal{E} \) and \( r \).

Subtract equation (1) from (2) to eliminate \( \mathcal{E} \):

\[ (3.42r - 1.47r) = (10.50 - 8.70) \] \[ 1.95r = 1.80 \]

Now, solve for \( r \):

\[ r = \frac{1.80}{1.95} \] \[ r \approx 0.923 \, \Omega \]

Now that we have the value for the internal resistance, we can find \( \mathcal{E} \) using either equation (1) or (2). Let's use equation (1):

\[ 1.47r = \mathcal{E} - 8.70 \] \[ 1.47 \times 0.923 = \mathcal{E} - 8.70 \] \[ 1.35641 = \mathcal{E} - 8.70 \] \[ \mathcal{E} = 1.35641 + 8.70 \] \[ \mathcal{E} \approx 10.05641 \, V \]

So the internal resistance of the battery \( r \approx 0.923 \, \Omega \) and the electromotive force of the battery \( \mathcal{E} \approx 10.06 \, V \).