Here, current, `I = 0.5 A` and time, `t = 1` hour `= 60 xx 60 s= 3600 s`
Thus, charge, `Q = I xx t = (0.5 A) (3600 s) = 1800 C`
Note that the voltage, i.e., `200 V` does not enter into calculation of charge flowing through the electric iron as it only maintains a current of `0.5 A` through the electric iron.