A battery of emf `1.4 V` and internal resistance `2 Omega` is connected to a resistor of `100 omega` resistance through an ammeter. This resistance of the ammeter is `4//3 Omega`. A voltmeter has also been connected to find the potential difference across the resistor.
a. Draw the circuit diagram.
b. The ammeter reads `0.02 A`. What is the resistance of the voltmeter?
c. The voltmeter reads `1.1 V`. What is the error in the reading?

Answer

(i) The circuit shows in the diagram
`(##BMS_V03_C06_E01_026_A01##)`
(ii) `200 Omega`
(iii) `-0.233 V`

Explanation

The circuit diagram is as shows.

(ii) Let resistance of the voltmeter be `R` ohm. The equivalent resistance of voltmetre (R ohm) and `100 Omega` in parellel is
`(100 xx R)/(100 + R) = (100R)/(100 + R)`
Resistance of the ammeter is `4//3 Omega`. Total resistance of the circuit is
`(100R)/(100 + R) + (4)/(3) + 2 Omega`
Current in the circuit as read by the ammeter is `0.02 A`.
Now, `0.02 = (1.4)/((100R)/(100 + R)+ (4)/(3) + 2)` `( :. I = (V)/(R))`
or `R = 200 Omega`
Therefore, resistance of the voltmeter is `200 Omega`.
(iii) Effective resistance between `B` and `C` is
`(100 xx 200)/(100 + 200) = (200)/(3) Omega`
The potential drop across this resistance is
circuit current `xx (200)/(3) = 0.02 xx (200)/(3) = (4)/(3) V = 1.333 V`
Reading of the voltmeter is `1.1 V`. Error in the reading of the voltmeter is `1.1 - 1.333 = - 0.233 V`.