Question

A voltmeter and an ammeter are connected in series across an ideal cell. When a certain resistance is connected in parallel to voltmeter, reading of ammeter increases three times and reading voltmeter reduces to one third. The ratio of resistance of voltmeter to that of ammeter is
A. 3
B. 1
C. 2
D. 5

Answer 1

Hint: The sum of voltage across ammeter and voltmeter is equal to the emf of the cell. Write the same equation when the resistor is connected across it. Solve the equations for voltage across ammeter and voltage across voltmeter. Use Ohm’s law to write the voltage in terms of resistance of the voltmeter and ammeter.

Formula used:
Ohm’s law is,
\[V = IR\]
Here, V is the voltage, I is the current and R is the resistance.


Complete step by step answer:
We have given that the voltmeter and ammeter are connected in the series with the ideal cell of potential difference E. Therefore, the voltage supplied by the ideal cell is equal to the voltage across ammeter plus the voltage across voltmeter.
\[{V_A} + {V_V} = E\] …… (1)
Now, the resistance is connected across the voltmeter. We can see the current will now also flow through the resistance. We have given that the reading of the ammeter increases to three times and voltage across voltmeter reduces to one third. Therefore,
\[3{V_A} + \dfrac{{{V_V}}}{3} = E\]
\[ \Rightarrow 9{V_A} + {V_V} = 3E\] …… (2)
From equation (1), we can substitute \[{V_V} = E - {V_A}\] in the above equation.
\[9{V_A} + E - {V_A} = 3E\]
\[ \Rightarrow 8{V_A} = 2E\]
\[ \Rightarrow {V_A} = \dfrac{E}{4}\] …… (3)
We substitute \[{V_A} = \dfrac{E}{4}\] in equation (2).
\[\dfrac{{9E}}{4} + {V_V} = 3E\]
\[ \Rightarrow {V_V} = 3E - \dfrac{{9E}}{4}\]
\[ \Rightarrow {V_V} = \dfrac{3}{4}E\] …… (4)
Now, using Ohm’s law, we can express the voltage across ammeter and voltmeter as follows,
\[{V_A} = I{R_A}\] …… (5)
 And,
\[{V_V} = I{R_V}\] …… (6)
Divide equation (4) by equation (3).
\[\dfrac{{{V_V}}}{{{V_A}}} = \dfrac{{\dfrac{3}{4}E}}{{\dfrac{E}{4}}}\]
\[ \Rightarrow \dfrac{{{V_V}}}{{{V_A}}} = 3\]
Using equation (5) and (6), we can write the above equation as,
\[\dfrac{{I{R_V}}}{{I{R_A}}} = 3\]
\[ \Rightarrow \dfrac{{{R_V}}}{{{R_A}}} = 3\]
Therefore, the ratio of resistance across voltmeter and ammeter is 3.

So, the correct answer is option (A).

Note:
In the second case we have given that reading in the ammeter increases to three times of the initial value. The reading in the ammeter is obviously the current in the circuit. Since the current increases to three times, the voltage across it also increases to three times as the resistance is constant. Students can answer almost every question relating voltage and current using Ohm’s law.