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**Hint:**Electric potential could be a location-dependent amount that expresses the number of potential energy per unit of charge at such location. Once a Coulomb of charge (or any given quantity of charge) possesses a comparatively great quantity of potential energy at a given location, then that location is alleged to be a location of high electric potential. And equally, if a Coulomb of charge (or any given quantity of charge) possesses a comparatively tiny amount of potential energy at a given location, then that location is alleged to be a location of low electric potential.

**Formula used:**

Heat produced,

$ \Rightarrow H = \dfrac{{{V^2}t}}{R}$

Where, $H$is the heat produced, $V$is the voltage, $t$is the time, and $R$is the resistance.

Voltmeter,

$ \Rightarrow V = R - IR$

Where, $R$ is the resistance, and $I$ is the ammeter.

**Complete step by Step Solution**In this question, the heat is altered and that causes the heat produced by is $9$ times its initial.

So they are asking the factor for which the applied potential difference changes.

$\left( a \right)$. Suppose the original potential difference applied to be $V$ and the original heat produced will be $H$.

Then,

$ \Rightarrow H = \dfrac{{{V^2}t}}{R}$

Now the new potential difference will be $V'$ and the heat produced after the change is $H'$

Then, we can write the equation as

$ \Rightarrow H' = \dfrac{{{{V'}^2}t}}{R}$

According to the question statement,

$ \Rightarrow H' = 9H$

Now we will put the values of both the heat,

We get

$ \Rightarrow \dfrac{{{{V'}^2}t}}{R} = 9 \times \dfrac{{{V^2}t}}{R}$

On further solving this equation, we get

$ \Rightarrow {V'^2} = 9{V^2}$

Which implies,

$ \Rightarrow V' = 3V$

Therefore we can say that the potential difference is increased by a factor $3$.

$\left( b \right)$. Let us consider the voltmeter and the ammeter to be in an ideal state.

Here the total resistance $R$ will be equal to

$ \Rightarrow R = 4 + 2$

$ \Rightarrow 6\Omega $

Then the ammeter reading will be,

$ \Rightarrow I = \dfrac{V}{R}$

$ \Rightarrow I = \dfrac{{12}}{6}$

$ \Rightarrow 2A$

Hence the ammeter reading will be$2A$.

Now we will calculate the voltmeter required,

$ \Rightarrow V = R - IR$

Substituting the values, we get

$ \Rightarrow 12 - 2 \times 2$

$ \Rightarrow 12 - 4$

$ \Rightarrow 8V$

**Therefore the voltmeter reading will be $8V$.**

**Note:**Voltmeters and ammeters measure the voltage and current, severally, of a circuit. Some meters in automobile dashboards, digital cameras, cell phones, and tuner-amplifiers are voltmeters or ammeters.

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