Question

Consider the diagram shown below. A voltmeter of resistance $150\Omega$ is connected across $A$ and $B$ . The potential drop across $B$ and $C$ measured by voltmeter is

$\left( a \right)\,\,29V$
$\left( b \right)\,\,27V$
$\left( c \right)\,\,31V$
$\left( d \right)\,\,30V$

Answer 1

A voltmeter is a device which is used for the measuring electric potential difference between any two points in an electrical circuit. Simple voltmeters move a pointer over a scale in relation to the voltage of the circuit; advanced voltmeters give a mathematical showcase of voltage by utilization of a simple to digital converter. The sum of voltage across resistors in series is equal to the voltage applied by the source connected.

Complete step by step solution:
When a voltmeter of the resistance $150\Omega$ is connected across the point $A$ and $B$ , the two resistance will be in parallel and the resistance across the point $A$ and $B$ will be
$\Rightarrow {R_{ab}} = \dfrac{{150 \times 100}}{{150 + 100}}$
After solving the above equation, we get
$\Rightarrow {R_{ab}} = 60\Omega$
Now the equivalent resistance of the circuit will be
$\Rightarrow {R_{ac}} = {R_{ab}} + R{}_{bc}$
After substituting the values of ${R_{ab}}$ and $R{}_{bc}$ the equation becomes
$\Rightarrow {R_{ac}} = 60 + 100$
And on solving it, we get
$\Rightarrow {R_{ac}} = 160\Omega$
The current in the circuit will be taken as
$\Rightarrow I = \dfrac{{50}}{{160}}$
After calculating we get
$\Rightarrow I = 0.31A$
In this way, Potential drop across B and C
$\Rightarrow {V_{bc}} = I{R_{bc}}$
After substituting the values in the above equation
$\Rightarrow {V_{bc}} = 0.31 \times 100$
Now we have to calculate the above equation we get
$\Rightarrow {V_{bc}} = 31V$

Hence the correct option is $\left( c \right)$ .

Note: Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Voltmeter uses the concept of potential gradient which is the fall in potential per unit length. So, when current passes through a wire which is of uniform area then the potential difference across any part is directly proportional to length of that path. If the potential gradient is small, then the sensitivity of voltmeter will be higher.