A circuit contains a 6V battery with internal resistance 2 ohms. Resistors P=10 ohms, Q=10 ohms, and R=15 ohms are connected as follows: P and R are in parallel, and this combination is in series with

Resistor Q is in series with the parallel combination of P and R.

This is the ammeter reading.

The ammeter reading is $\boxed{\text{0.33 A}}$.

11.2.2

The switch S is now open.

The external circuit now consists only of resistor P (10 $\Omega$) in series with resistor Q (10 $\Omega$).
The new total external resistance ($R'_{ext}$) is:

The new total resistance of the circuit ($R'_{total}$) is:

The total resistance has increased.

The new total current ($I'_{total}$) would be:

Since $0.27 \, \text{A} < 0.33 \, \text{A}$, the ammeter reading will DECREASE.

When switch S is opened, resistor R is removed from the parallel branch. This increases the total external resistance of the circuit. According to Ohm's law, an increase in total resistance leads to a decrease in the total current flowing through the circuit.
The ammeter reading will $\boxed{\text{DECREASE}}$.

11.2.3

The voltmeter measures the terminal potential difference ($V_{terminal} = \mathcal{E} - Ir$).
From 11.2.2, we determined that when switch S is opened, the total current ($I$) in the circuit decreases.
Since the EMF ($\mathcal{E}$) and internal resistance ($r$) are constant, a decrease in $I$ means that the voltage drop across the internal resistance ($Ir$) will decrease.
Therefore, $V_{terminal} = \mathcal{E} - Ir$ will increase because a smaller value is subtracted from the constant EMF.
The voltmeter reading will $\boxed{\text{INCREASE}}$.

11.2.4

The voltmeter measures the terminal potential difference, which is given by the formula $V_{terminal} = \mathcal{E} - Ir$. When switch S is opened, the total external resistance of the circuit increases, causing the total current ($I$) to decrease. Since the internal resistance ($r$) is constant, a decrease in the total current ($I$) results in a smaller voltage drop ($Ir$) across the internal resistance. Consequently, the terminal potential difference ($V_{terminal}$) increases because a smaller amount is subtracted from the constant EMF ($\mathcal{E}$).