(a) For the combination of resistors shown in the following figure, find the equivalent resistance between M and N. (b) State Joule.s law of heating. (c ) Why are need a 5A 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 heaterin series?

(a) For the combination of resistors shown in the following figure, fi

1.	(a) For the combination of resistors shown in the following figure, find the equivalent resistance between M and N. (b) State Joule.s law of heating. (c ) Why are need a 5A 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 heaterin series?
Answer» Solution :(a) We have `R_(1) and R_(2)` is series with the parallel combination of `R_(3) and R_(4)`. Hence, equivalent resistance between M and N is : `R=R_(1)+R_(2)+(R_(3)R_(4))/((R_(3)+R_(4)))` (b) When a current I is PASSED through a resistor R for a time t, the amount of heat produced .H. is given by the RELATION `H=I^(2)Rt` This is known as Joule.s law. (c ) For an electric iron consuming 1 kW electric power at 220 V, current FLOWING through the element. `I=(P)/(V)=(1kW)/(220V)=4.54A` Therefore, we use a fuse of 5 A current rating. (d) In a series circuit same current flows throughout the electric circuit. Thus, it is impracticable to connect an electric bulb and an electric heater in series because they need CURRENTS of widely different values to operate properly.