CBSE Class 10 Science 2019 · Paper 31/1/2/C · Question 14

(a) Total resistance of the network $=3\Omega$ (each branch is $6\Omega$, and two $6\Omega$ branches in parallel give $\frac{6\times 6}{6+6}=3\Omega$). (b) Current through the ammeter $=\frac{6\text{V}}{3\Omega }=2$ A. (c) Each branch carries $1$ A, so the potential difference across each $3\Omega$ resistor is $1\times 3=3$ V, and across the $4\Omega$ resistor it is $1\times 4=4$ V.