(a) For the combination of resistors shown in the following figure, find the equivalent resistance between M & N.(b) State Joule’s law of heating.(c) Why we need a 5 A fuse for an electric iron which consumes 1 kW power at 220 V?(d) Why is it impracticable to connect an electric bulb and an electric heater in series?"

(a) Given:

 Resistance of the resistors = $R_1, R_2, R_3,\ and\ R_4$ 

To find: Equivalent resistance, $(R_eq)$ between M & N.

Solution:

Here, from the figure we can see that $R_3$ and $R_4$ are in parallel combination.

We know that  when the resistance of two conductors are connected in parallel, then the total resistance is given as-

$\frac {1}{R_p}=\frac {1}{R_3}+\frac {1}{R_4}$

$\frac {1}{R_p}=\frac {R_4+R_3}{R_3\times {R_4}}$

${R_p}=\frac {R_3\times {R_4}}{R_4+R_3}$

Now, $R_1$, $R_2$ and $R_p$ are in series.

Then the equivalent resistance is given as-

$R_eq=R_1+R_2+R_p$

$R_eq=R_1+R_2+\frac {R_3\times {R_4}}{R_4+R_3}$

Thus, the equivalent resistance, $(R_eq)$ between M & N is $R_1+R_2+\frac {R_3\times {R_4}}{R_4+R_3}$.

(b) Joule's Law of Heating refers to the heat generated by an electrical conductor is directly proportional to the product of its resistance $(R)$ and the square of the electric current passing through the conductor with time.

Mathematical expression for Joule's law of heating is:

$H=I^2Rt$

Where,

$H$ = Heating effect.

$I$ =  Current flowing through the conductor.

$R$ = Amount of electric resistance in the conductor.

$t$ = Time taken.

(c) Given:

Voltage, $V$ = 220 V

Power, $P$ = 1 kW = 1000 W

Fuse current, $I$ = 5 A

To find: Current flow in the circuit, $I$.

Solution:

We know that, power is given as-

$P=V\times {I}$

Substituting the given values we get-

$1000=220\times {I}$

$I=1000\times {220}$

$I=4.54A$

Thus, the current flow in the circuit is 4.54A. 

We should use a fuse with a current capacity slightly greater than the maximum current drawn by the device (here, electric iron). Thus, a 5 A fuse is needed, as it is slightly greater than 4.54 A. Now, if the value of current exceeds 5 A, the fuse wire will melt.

(d) It is impracticable to connect an electric bulb and an electric heater in series because they will not get currents and voltages as per their requirement.