Question

Figure shows two resistors ${R_1}$ and ${R_2}$ connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of 300ohm is used to measure potential difference across ${R_1}$. Find the reading of the voltmeter?

Answer 1

Hint: we will first acknowledge the formula for current. Then we will notice the resistors connected in parallel and the resistors connected in series. We will apply the formulas for the same respectively and then solve this answer further. Refer to the solution below.
Formula used: $I = \dfrac{V}{{{R_{eq}}}}$
If the resistors are connected in parallel then,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} - - - \dfrac{1}{{{R_n}}}$
If the resistors are connected in series then,
 ${R_{eq}} = {R_1} + {R_2} - - - {R_n}$


Complete step-by-step answer:
As per given in the question, the current in the circuit is given by $I$.
The formula for which is-
$ \Rightarrow I = \dfrac{V}{{{R_{eq}}}}$.
The resistance provided by the voltmeter as per given in the question is- ${R_v} = 300\Omega $.
The value of ${R_1}$ given in the question is- ${R_1} = 200\Omega $.
The value of ${R_2}$ as per given in the question is- ${R_2} = 880\Omega $.
The formula for resistors connected in parallel is always calculated as-
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} - - - \dfrac{1}{{{R_n}}}$
Now, as we can see from the figure, the resistance provided by the voltmeter and the resistance ${R_1}$ are in parallel. Let their equivalent be ${R_3}$. Thus-
$
   \Rightarrow \dfrac{1}{{{R_3}}} = \dfrac{1}{{{R_v}}} + \dfrac{1}{{{R_1}}} \\
    \\
   \Rightarrow \dfrac{1}{{{R_3}}} = \dfrac{1}{{300}} + \dfrac{1}{{200}} \\
    \\
   \Rightarrow \dfrac{1}{{{R_3}}} = \dfrac{{200 + 300}}{{300 \times 200}} \\
    \\
   \Rightarrow {R_3} = \dfrac{{60000}}{{500}} \\
    \\
   \Rightarrow {R_3} = \dfrac{{600}}{5} \\
    \\
   \Rightarrow {R_3} = 120\Omega \\
$
The formula for resistors connected in series is-
$ \Rightarrow {R_{eq}} = {R_1} + {R_2} - - - {R_n}$
Now, as we can see from the figure, ${R_3}$ and ${R_2}$ are in series. Let their equivalent be ${R_{eq}}$. Thus-
$
   \Rightarrow {R_{eq}} = {R_3} + {R_2} \\
    \\
   \Rightarrow {R_{eq}} = 120 + 880 \\
    \\
   \Rightarrow {R_{eq}} = 1000\Omega \\
$
Since we already knew that the formula of current is given by $I = \dfrac{V}{{{R_{eq}}}}$. Hence-
$
   \Rightarrow I = \dfrac{V}{{{R_{eq}}}} \\
    \\
   \Rightarrow I = \dfrac{{40}}{{1000}} \\
    \\
   \Rightarrow I = 0.04A \\
$
Now, the reading of the voltmeter will be equal to the potential difference. Hence-
$
   \Rightarrow I = \dfrac{V}{{{R_{eq}}}} \\
    \\
   \Rightarrow V = I \times {R_{eq}} \\
    \\
   \Rightarrow V = 0.04 \times 120 \\
    \\
   \Rightarrow V = 4.8volt \\
$


Note: A voltmeter is an electric potential difference measurement device used in an electrical circuit to compare between two points. A scale is adjusted by a traditional analog voltmeter in relation to the voltage of the circuit; a graphical representation of the voltage by an analog to digital converter is provided by digital voltmeters.