CBSE Class 10 Science 2019 · Paper 31/5/1 · Question 19(a)

Circuit A (series): total resistance $R=1+2=3\Omega$, so a current $I=\frac{V}{R}=\frac{6}{3}=2$ A flows through both resistors. Power used in the $2\Omega$ resistor: $P_A=I^{2}R=2^2\times 2=8$ W. Circuit B (parallel): the $2\Omega$ resistor has the full $4$ V across it, so $P_B=\frac{V^2}{R}=\frac{4^2}{2}=8$ W. The power used in the $2\Omega$ resistor is the same, $8$ W, in both circuits: $P_A:P_B=1:1$.