(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) The resistance R3 and R4 are in parallel combination with each other and the combination is in series resistance R2 and R1.
The effective value of R3 and R4 is,

Equivalent set between M and N

(b) Joule’s law of heating states that when a current ‘I’ passes through a conductor of resistance ‘R’ for time ‘t’ then the heat developed in the conductor is equal to the product of the square of the current, the resistance and time.
H = I2Rt
(c) In the given case the maximum current drawn by the electric iron is
P = V I
I = P/V
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This value is less than the current rating of the fuse. Hence, a 5 A rated fuse can be used.
(d) When a bulb and a heater are connected in series the current remains the same and voltage is divided according to the resistance of the appliances. Now, the reduction in the voltage ultimately reduces the power delivered to the devices.
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Where V is voltage and R is resistance
if V reduces one fold, power will reduce by 2 times.
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