Question

If the voltmeter reads 0.21V and the ammeter reads 0.1A, the resistance of the ammeter is (in ohm)

A) 0.01
B) 0.15
C) 0.2
D) 0.1

Answer 1

Hint
In this question as ammeter and resistance are kept in series so their resistances will add and after that we will apply the ohm’s law i.e. $V = IR$, I is the current and R is the total resistance. On substituting the values, we will get the desired result.

Complete step by step answer
As in the circuit diagram, resistance $2\Omega $ and ammeter is connected in series and we have to calculate the value of the resistance of the ammeter.
For this, let us consider RA is the resistance of the ammeter which we have to calculate.
Now, as the resistance$2\Omega $ and ammeter is connected in series then we can write the total resistance which is the sum of the both the resistances, i.e.
$R = 2\Omega + {R_A}$ ……………………… (1)
Now, as it is given that
Voltage applied across the circuit is $V = 0.21V$
And the current passing through the circuit is $I = 0.1A$
Now, applying the ohm’s law which states that voltage is proportional to the current and the current.
$ \Rightarrow V = IR$
Where R is the total resistance of the circuit
Now put the all values in above equation, we get
$\Rightarrow 0.21 = 0.1 \times \left( {{R_A} + 2} \right) $
$\Rightarrow \dfrac{{0.21}}{{0.1}} = {R_A} + 2 $
$\Rightarrow {R_A} = 2.1 - 2 $
$\Rightarrow {R_A} = 0.1\Omega $
Hence, the resistance of the ammeter is 0.1 ohm.
Thus, option (D) is correct.

Note
It must be notice that when two resistances are connected in the series their resistance increases, it means the total resistance is the sum of all the resistances i.e. $R = {R_1} + {R_2} + {R_3} +. ...........$
But when the resistances are connected in parallel arrangement then their total resistance decreases which can be written as $\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} +. ............$
Therefore, care must be taken in calculating the total resistance of the circuit.