In the figure shown below, a resistance of 150.4 $$\Omega$$ is connected in series to an ammeter A of resistance 240 $$\Omega$$. A shunt resistance of 10 $$\Omega$$ is connected in parallel with the ammeter. The reading of the ammeter is _________ mA.

Given:
ammeter resistance = 240 Ω
shunt = 10 Ω (in parallel with ammeter)
series resistor = 150.4 Ω
battery = 20 V

step 1: equivalent of ammeter branch

ammeter ∥ shunt:

R₁ = (240 × 10) / (240 + 10)
= 2400 / 250
= 9.6 Ω

step 2: total circuit resistance

$$R_{total}=150.4+9.6=160Ω$$

step 3: total current

$$I_{total}=20/160=0.125A$$

step 4: current division in parallel branch

current through ammeter:

$$I_A=I_{total}\times\frac{shunt}{(ammeter+shunt)}$$
= 0.125 × (10 / 250)
= 0.125 × 0.04
= 0.005 A

=5 mA