A cell of emf `3.4 V` and internal resistance `3 Omega` is connected to an ammeter having resistance `2Omega` and to an external resistance of `100Omega`. When a voltmeter is connected across the `100 Omega` resistance, the ammeter reading is `0.04 A`. Find the voltage reading by the voltmeter and its resistance. Had the voltmeter been an ideal one what would have been its reading?

Options

A. `400 Omega`
B. `200 Omega`
C. `300 Omega`
D. `500 Omega`

Answer

A

Explanation

Let `R_(1)` and `R_(2)` be the resistances of the ammeter and the voltmeter, respectively. Let the external resistance be denoted by `R` and the internal resistance of battery be `r`. The equivalent resistance of the parallel combination of `R` and `R_(2)` is given by
`R' = (R R_(2))/(R + R_(2))`
The total resistance `RT` of circuit then becomes
`RT = R_(1) + r + (R R_(2))/(R + R_(2))`
The current in the circuit is given by
`I = (E)/(R_(1) + r+ (R R_(2))/(R + R_(2)))`
This must be equal to `0.04 A`, the reading indicated by the ammeter.
`(3.4)/(2 + 3+ (100R_(2))/(100 + R_(2))) = 0.04` or `R_(2) = 400 Omega`
In case of an ideal voltmeter, no current flows through it. In that case, current in the circuit is
`I' = (3.4)/(2 + 3 + 100) = (3.4)/(105) = 0.0324 A`
Potential drop across the resistance `R` would br `100 xx 0.0324 = 3.24 V`.
This should be that reading indicated by an ideal voltmeter.