A perspex ruler is rubbed with a silk cloth, what charge will the ruler obtain? Resistors in series are __ dividers. An object that opposes the flow of current is a . Voltmeter measures the amount of

$V_5$ measures the total voltage across the battery pack.

There are three $1.5 \text{ V}$ cells connected in series.

The reading on $V_5$ is $\boxed{4.5 \text{ V}}$.

The $2 \Omega$ resistor and the unknown resistor are in parallel. This parallel combination is in series with the $1 \Omega$ resistor (under $V_4$). The ammeter A is in the main circuit. $V_1$ measures the voltage across the $2 \Omega$ resistor.

The reading on $V_4$ is $3 \text{ V}$ and the resistance under $V_4$ is $1 \Omega$.
Using Ohm's Law $V = IR$:

This is the total current flowing through the series part of the circuit.

The total voltage supplied by the cells is $4.5 \text{ V}$.
The voltage drop across the $1 \Omega$ resistor is $V_4 = 3 \text{ V}$.
The voltage across the parallel combination ($V_1$ and $V_2$) is the total voltage minus the voltage drop across $R_4$.

Since $V_1$ measures the voltage across the $2 \Omega$ resistor, which is part of the parallel combination, and voltage across parallel components is the same:

The reading on $V_1$ is $\boxed{1.5 \text{ V}}$.

Ammeter A is in the main circuit, measuring the total current.

The current flowing through the $1 \Omega$ resistor (under $V_4$) is the total current in the circuit because it is in series with the parallel combination.

The reading on A is $\boxed{3 \text{ A}}$.

The combined parallel resistance is $1.5 \Omega$. Let the unknown resistor be $R_x$.
The $2 \Omega$ resistor is in parallel with $R_x$.
The formula for two parallel resistors is:

The resistance of the unknown resistor is $\boxed{6 \Omega}$.

Current $I$ is the rate of flow of charge $Q$ over time $t$:

The current through the ammeter is $I = 3 \text{ A}$.
The time is $t = 2 \text{ s}$.

The amount of charge that flows through the ammeter in 2 seconds is $\boxed{6 \text{ C}}$.

The resistor under $V_4$ has a resistance $R_4 = 1 \Omega$.
The voltage across it is $V_4 = 3 \text{ V}$.
The current through it is $I = 3 \text{ A}$ (from part ii).
The time is not specified for this part, but typically for energy transfer questions related to heating, a time duration is needed. Assuming the question refers to the same time duration as part (v), which is $t = 2 \text{ s}$.

The energy transferred (heat) $E$ can be calculated using the formula:

or

or

Using $V_4 = 3 \text{ V}$, $I = 3 \text{ A}$, and $t = 2 \text{ s}$:

The amount of energy transferred in the resistor is $\boxed{18 \text{ J}}$.