Question

An ammeter has range ${\text{I}}$ with resistance ${R_0}$, which of the following resistance can be connected in series with it to decrease its range to $\dfrac{I}{n}$.
(A) $\dfrac{R_0}{n}$
(B) ${R_0}(n-1)$
(C) $\dfrac{R_0}{(n+1)}$
(D) None of these

Answer 1

Hint
Potential difference across the ends of ammeter always remains the same; we only need to change the resistance of the ammeter. If voltage is constant current is inversely proportional to value of resistance. Then, calculate the value of resistance for which current reduces by $n$ times.

Complete step by step solution
Given, current through ammeter ${\text{I}}$ when resistance of ammeter is ${R_0}$.
Let potential difference across ends of resistance is $V$.
We know that potential difference across ammeters remains the same for a particular circuit, if we change the resistance if ammeter it just changes the range of ammeter.
Let we connect a resistance $R$ in series with ammeter.
We know, \[{\text{Potential Difference = Current}} \times {\text{Resistance}}\].
When the range of ammeter is ${\text{I}}$,
$V = I \times {R_0} … (1)$
When the range of ammeter is $\dfrac{I}{n}$,
$V = \dfrac{I}{n} \times (R + {R_0}) ... (2)$
From equation (1) and (2), we have
$\dfrac{I}{n} \times (R + {R_0}) = I \times {R_0}$
After solving above equation,
$R = {R_0}(n - 1)$.
Hence the correct answer is option (B).

Note
Ammeters are made by galvanometer by connecting a short resistance in parallel with a galvanometer, and the range of ammeters is dependent on this short resistance. To change the range of ammeters just modify the short resistance. To convert a galvanometer into voltmeter a large resistance is connected in series with the galvanometer.