A potential differnce of 250 volts is applied across a resistance of 1000 `Omega` in an electric iron. Find (1) the current (2) the heat produced in joule in 12 seconds.
keeping other conditions the same , if the length of the wire in the iron is reduced to half the original length (by cutting the wire ), what will be the current and heat produced ?

Explanation

Data : `V=250 V, R=1000 Omega , t=12s , I= ? , H=?`
(1) `V=IR therefore I=V/R =(250V)/(1000 Omega)=0.25A`
The current through the resistance =0.25 A
(2) `H=I^(2)Rt`
`=(0.25A)^(2)xx1000 Omega xx 12 `s
`=((1)/(4) xx 1000)xx(1/4 xx 12) J`
`=250xx3J =750J`
Or `H=VIt =250 V xx 0.25 A xx12s`
`=250xx3J=750 J`
The heat energy produced in the resistance in 12 seconds =750 joules .
On cutting the wire , the resistance of the wire will become half the initial resistance . hence, the current will become double the initial current as I=V/R and V is the same in both the cases. Therefore, the current in the wire will be `0.25Axx2=0.5A` (Hence , the heat produced will be VIt =250 V `xx0.5 A xx 12s =250 xx6J=1500 J`.