CBSE Class 10 Science 2020 · Paper 31/3/1 · Question 29

(a) $R_3$ and $R_4$ are in parallel: $R_p=\frac{R_3{R}_4}{R_3+R_4}$. This is in series with $R_1$ and $R_2$, so $R_{\text{eq}}=R_1+R_2+\frac{R_3{R}_4}{R_3+R_4}$. (b) Joule's law of heating: the heat produced in a resistor is directly proportional to the square of the current for a given resistance, directly proportional to the resistance for a given current, and directly proportional to the time for which the current flows: $H=I^{2}Rt$. (c) In normal use the iron draws $I=\frac{P}{V}=\frac{1000}{220}\approx 4.55$ A. A $5$ A fuse carries this current without melting, but melts if the current rises above $5$ A, so it protects the iron and the circuit; hence a $5$ A fuse is needed. (d) In series the same current flows through both, but a bulb and a heater need currents of widely different values, so neither would get the current and voltage it needs (the heater would hardly heat); also, if one of them fails, the circuit breaks and the other stops working too.