A cell of emf `2V` and internal resistance `0.1 Omega` supplies a current through a coil of resistance `11.0 Omega`. The current is being measured by an ammeter whose resistance is `6 Omega`. What reading does it give ? What is the percentage difference from the actual current, when the meter is not used ?

Answer

`0.11A; 33.3 %`

Explanation

When ammeter is in the circuit total resistance of circuit `= 0.1+ 11.9 xx 6 = 18 Omega` current in the circuit, `I = 2//18 = 1//9 = 0.11A`. When ammeter in not in the circuit total resistance of the circuit `= 0.1 = 11.9 = 12 Omega`
Current is the circuit, `I' = 2//12 = 1//6 A`
% difference in circuit `= (I'- I)/(I') xx 100`
`= ((1//6) - (1//9))/((1//6))xx 100 = 33.3 %`