CBSE Class 10 Science 2018 · Paper 31/1/C · Question 12(a)

Ammeter reading: $0.24$ A. Voltmeter reading: $2.88$ V. Reason: With the key closed, the $5\Omega$, $8\Omega$ and $12\Omega$ resistors, the ammeter and the battery form one loop, i.e. they are in series. So the total resistance is $R=5+8+12=25\Omega$, and the same current $I=\frac{V}{R}=\frac{6\text{V}}{25\Omega }=0.24$ A flows through every part; the ammeter, being in series, reads this current. The voltmeter is connected in parallel across the $12\Omega$ resistor only, so it reads the potential difference across that resistor: $V=IR=0.24\text{A}\times 12\Omega =2.88$ V (the battery and the meters being taken as ideal).