physics question electricity

When switch S is closed, the voltmeter reading decreases and the ammeter reading increases as you said.
When the switch is closed, the voltage drops across the internal resistance, which is known as lost volts. As the terminal potential difference decreases, the lost volts increases.
As a reminder: When 2 resistors are connected in parallel, the voltage across both resistors is the same, but the battery current is split between the two resistors. The resistor with the smaller resistance has the largest current through it.

hi, please could I have some help on this question? I know that the ammeter reading will increase because total resistance will decrease but why does V decrease? Surely it sould increase if current increases?
here is the question: https://ibb.co/y4Cw6dY
thanks!

You used the correct relationship or equation in post #3 or reply #3 but explained it incorrectly.
As you know the (total) current in the circuit or the battery increases and using the relationship
terminal p.d. = emf - p.d. across the internal resistor
OR
V = E - Ir
we can see that the p.d. across the internal resistor increases (because current increases) and the terminal p.d. will decrease.
The voltmeter is measuring the terminal p.d. .
Emf of the battery usually does not change in A level exam or most of the univ physics course.

oh, so since I in V=E-Ir is increasing (which is the current in the internal resistor) V which is the terminal pd will decrease?

Indeed, the terminal p.d. will decrease when the current increases.