CBSE Class 10 Science 2020 · Paper 31/5/1 · Question 29

(a) Each lamp gets the full 220 V, so $I_1=\frac{100\text{W}}{220\text{V}}=\frac{10}{22}$ A and $I_2=\frac{10\text{W}}{220\text{V}}=\frac{1}{22}$ A; the total current is $I=I_1+I_2=\frac{11}{22}$ A $=0.5$ A. (b) (i) In parallel, X ($2\Omega$) and Y ($3\Omega$) are connected between the same two junctions, which are joined through a plug key to the 5 V battery; in series, X and Y are joined end to end in a single loop with the plug key and the 5 V battery (circuit diagrams in the figure). (ii) In series, $R_s=2\Omega +3\Omega =5\Omega$, so $I=\frac{5\text{V}}{5\Omega }=1$ A, and the voltage across the $3\Omega$ resistor is $V=IR=1\text{A}\times 3\Omega =3$ V.