Question

A potential divider consists of two resistors of resistances \[{{R}_{1}}\] and \[{{R}_{2}}\]connected in series across a source of potential difference \[{{V}_{in}}\]. The potential difference across \[{{R}_{1}}\] is \[{{V}_{out}}\]. Which charges to \[{{R}_{1}}\And {{R}_{2}}\]will increase the value of \[{{V}_{out}}\]?

\[{{R}_{1}}\]	\[{{R}_{2}}\]
A- Doubled	Doubled
B- Doubled	Halved
C- Halved	Doubled
D- Halved	halved

Answer 1

Hint: A potential divider is a simple circuit that uses resistors or a combination of resistors to supply a variable potential difference as per the need of the output circuit. In this circuit both the given resistances are in series combination but the output is obtained only across the first resistor.

Complete step by step answer:
As per the given question, potential difference across the given resistor can be found by using ohm’s law. The total resistance of the circuit is \[{{R}_{1}}+{{R}_{2}}\] and the input voltage fed to the circuit is \[{{V}_{in}}\].
$ {{V}_{in}}=I({{R}_{1}}+{{R}_{2}}) \\ $
$\Rightarrow I=\dfrac{{{V}_{in}}}{{{R}_{1}}+{{R}_{2}}} \\ $
Now, again using ohm’s law to find the potential difference across first resistor,
$ V=I{{R}_{1}} \\ $
$ \Rightarrow V=\dfrac{{{V}_{in}}{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}} \\ $
We can see that the output voltage is directly proportional to the value of first resistor, thus we can write as
$ \Rightarrow {{V}_{out}}\propto {{R}_{1}} \\ $
$\therefore {{V}_{out}}\propto \dfrac{1}{{{R}_{1}}+{{R}_{2}}} \\ $

So, the correct answer is “Option B”.

Note:
Ohm’s law holds for all ohmic conductors and it is one of the most important laws in electricity and magnetism. The combination was in series so we directly added the two resistances, had they been in parallel combination we would have to add them in another way as we use to add the resistances in parallel combination.