Study the circuit and find out-

(a) Current in 12 ohm resistor

(b) Difference in the readings of ammeter A₁ and A₂ if any

(a) From the diagram,

Two resistors of 24 ohm are connected in parallel and their effective resistance is given by,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{\text R1} + \dfrac {1}{\text R_2}

Here, $\text R$1 = \text R2 = 24 ohm,

$\dfrac {1}{\text R$\text P} = \dfrac {1}{24} + \dfrac {1}{24} \\[1em] = \dfrac {2}{24} \\[1em] \Rightarrow \dfrac {1}{\text R\text P} = \dfrac {1}{12} \\[1em] \Rightarrow \text R_\text P = 12 \text { ohm}

Since $\text R_\text P$ and 12 ohm resistance are in series then their effective resistance is given by,

$\text R$\text S = \text R\text P + 12 \\[1em] = 12 + 12 \\[1em] \Rightarrow \text R_\text S = 24 \text { ohm}

Let net current flowing through the circuit is '$\text I$'.

Now, this same current flows through $\text R_\text P$ and 12 ohm resistance then

$\text I = \dfrac {\text {Voltage of battery}}{\text R_\text S} \\[1em] \dfrac {6}{24} \\[1em] \Rightarrow \text I = 0.25\ \text A$

Hence, the current flowing in 12 ohm resistor is 0.25 A.

(b) The total current flowing from the battery passes through A₁, A₂ and the resistors and finally returns to the battery. Since no current is lost or divided at any point, A₁ and A₂ will show the same reading of 0.25 A.

Then,

Difference in the readings of ammeter A₁ and A₂ = 0.25 - 0.25 = 0 A

Hence, the difference in the readings of ammeter A₁ and A₂ is zero.