Solveeit Logo

Question

Question: What is the resistance of the voltmeter shown in the circuit? ![](https://www.vedantu.com/quest...

What is the resistance of the voltmeter shown in the circuit?


A. 8003Ω\dfrac{{800}}{3}\Omega
B. 800Ω800\,\Omega
C. 1000Ω1000\,\Omega
D. 200Ω200\,\Omega

Explanation

Solution

In a series connection, the current is each element is equal to the supply current
I=I1=I2.......I = {I_1} = {I_2}....... whereas the supply voltage is equal to the sum of individual voltages
V=V1+V2.......V = {V_1} + {V_2}.......
Ohm’s law states that for given physical conditions, the voltage drop across a particular element in a circuit is directly proportional to the current flowing through the element.
The mathematical expression is given as VIV \propto I
Replacing the proportionality sign we get V=IRV = IR where R is the resistance of the element.

Complete step by step solution:
Since the voltmeter is connected in parallel across the 8003Ω\dfrac{{800}}{3}\Omega , the voltage drop across the 8003Ω\dfrac{{800}}{3}\Omega resistor is 50V50\,V .
We know that in a series connection the voltage is the sum of voltage drops across each element.
Hence, 100=V200Ω+V8003Ω100 = {V_{200\Omega }} + {V_{\dfrac{{800}}{3}\Omega }}
V200Ω=50V\Rightarrow {V_{200\Omega }} = 50V .
Now the current flowing through the 200Ω200\,\Omega is I=VRI = \dfrac{V}{R} (by ohm’s law)
I=50200\Rightarrow I = \dfrac{{50}}{{200}}
I=14A\Rightarrow I = \dfrac{1}{4}A
In a series connection the current flowing through each element is the same.
Hence, I(8003Ω+50V)=14A{I_{(\dfrac{{800}}{3}\Omega + 50V)}} = \dfrac{1}{4}A where 8003Ω+50V\dfrac{{800}}{3}\Omega + 50V represents the parallel connection of the 50V50\,Vvoltmeter and the 8003Ω\dfrac{{800}}{3}\Omega resistance.
The equivalent resistance of the connection( 50V50\,Vvoltmeter and the 8003Ω\dfrac{{800}}{3}\Omega resistance) is given as
R=VIR = \dfrac{V}{I} using ohm’s law
R=5014\Rightarrow R = \dfrac{{50}}{{\dfrac{1}{4}}}
R=200Ω\Rightarrow R = 200\Omega
So, the equivalent resistance of the 8003Ω+50V\dfrac{{800}}{3}\Omega + 50V combination is R=200ΩR = 200\Omega .
Let the resistance of the voltmeter be RV{R_V}.
The parallel equivalent is given by R=RV×8003RV+8003R = \dfrac{{{R_V} \times \dfrac{{800}}{3}}}{{{R_V} + \dfrac{{800}}{3}}}
We have calculated the value of parallel equivalent to be R=200ΩR = 200\Omega
Substituting in the equation
200=RV×8003RV+8003200 = \dfrac{{{R_V} \times \dfrac{{800}}{3}}}{{{R_V} + \dfrac{{800}}{3}}}
Further solving this
200=800RV3RV+800\Rightarrow 200 = \dfrac{{800{R_V}}}{{3{R_V} + 800}}
600RV+160000=800RV\Rightarrow 600{R_V} + 160000 = 800{R_V}
RV=800Ω\Rightarrow {R_V} = 800\Omega
Hence option B is correct.

Note: Voltmeters are devices with high resistances to allow least possible current to flow and are always connected in parallel to prevent heating of the circuit. Simplifying the circuit into smaller equivalents and then backward substituting the values is better than introducing new circuit variables for every branch in the circuit since it prevents the situation of handling multiple equations at the same time.