So, here for question no. 4 and 5, we are given a Diagram, which have 3 resistors connected in parallel each having a resistance of [tex]{\bf{2\Omega}}[/tex], [tex]{\bf{3\Omega}}[/tex] and [tex]{\bf{4\Omega}}[/tex]. And we are told to find a factor which will change for the given combination, and a factor which will not change, So here, as the resistors are connected in parallel, so the voltage will remain same for all the resistors, while the Current will be different for every resistor, And that's what the reason of why we always do connection in parallel for home purposes, because, we need voltage same in every resistor of the circuit. So, the factor which will change is [tex]{\boxed{\bf{Current}}}[/tex] and the factor which willn't change is [tex]{\boxed{\bf{Voltage}}}[/tex], and if the resistors were connected in Series combination, then current should be same for all resistors, while voltage will change.
Well, if the resistors are connected in the above way we have, and if we call our Resistors as [tex]{\bf{R_{1}}}[/tex], [tex]{\bf{R_{2}}}[/tex] and [tex]{\bf{R_{3}}}[/tex], then the equivalent resistance we will be having is :
[tex]{:\implies \quad \boxed{\bf{\dfrac{1}{R_{eq.}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}}}[/tex]
Also, we can prove this fact just with the help of the reason provided in the above question, i.e that in Parallel combination, voltage remains same, but current changes, while in Series combination, current remains same but voltage changes, so we are just gonna prove the formula for both series combination and Parallel as well
So, let's prove the formula for series combination first, so consider n resistors connected in series, let Resistors as [tex]{\bf{R_{1},R_{2},R_{3}\cdots \cdots R_{n}}}[/tex], now as current will be same through all the resistors as connected in series, so if the current [tex]{\bf{I}}[/tex] from then, same current will flow through the whole circuit and each resistors, also let the voltages of every resistors being [tex]{\bf{V_{1},V_{2},V_{3}\cdots \cdots V_{n}}}[/tex], so now by Ohm's law, we can deduce that, the current from first circuit will be [tex]{\bf{V_{1}=IR_{1}}}[/tex]...and so on for all Resistors, and now if we add all resistors, so the equivalent Voltage will just be ;
[tex]{:\implies \quad \sf V_{eq.}=V_{1}+V_{2}+V_{3}+\cdots \cdots +V_{n}}[/tex]
[tex]{:\implies \quad \sf IR_{eq.}=IR_{1}+IR_{2}+IR_{3}+\cdots \cdots IR_{n}}[/tex]
[tex]{:\implies \quad \sf I(R_{eq.})=I(R_{1}+R_{2}+R_{3}+\cdots \cdots R_{n})}[/tex]
[tex]{:\implies \quad \boxed{\bf{R_{eq.}=R_{1}+R_{2}+R_{3}+\cdots \cdots R_{n}}}}[/tex]
Therefore, we proved for series.
Now, consider n resistors connected in parallel, with resistances, [tex]{\bf{R_{1},R_{2},R_{3}\cdots \cdots R_{n}}}[/tex], now as voltage will be same through all the resistors as connected in parallel,while current will change, so let total voltage be [tex]{\bf{V}}[/tex], and current for each resistor as [tex]{\bf{I_{1},I_{2},I_{3}\cdots \cdots I_{n}}}[/tex],so by using the same concept as above we can write by using ohm's law :
[tex]{:\implies \quad \sf I_{eq.}=I_{1}+I_{2}+I_{3}+\cdots \cdots +I_{n}}[/tex]
Rewrite using ohm's law :
[tex]{:\implies \quad \sf \dfrac{V}{R_{eq.}}=\dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_3}+\cdots \cdots +\dfrac{V}{R_n}}[/tex]
[tex]{:\implies \quad \sf V\dfrac{1}{R_{eq.}}=V\bigg(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\cdots \cdots +\dfrac{1}{R_n}\bigg)}[/tex]
[tex]{:\implies \quad \boxed{\bf{\dfrac{1}{R_{eq.}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}+\cdots \cdots +\dfrac{1}{R_{n}}}}}[/tex]
Hence, we proved for Parallel combination too
Also, refer to the attachment for a better understanding, and just a point that Voltmeter is always connected in parallel while ammeter in series, and all the things above are taken in SI units
