i house, served by a 220 V supply line, is protected y a 9A fuse. What is the maximum number of 60 ratt bulbs that can be turned on in parallel?

To find the maximum number of bulbs, we first calculate the total power capacity of the circuit using the formula P = V × I. With a supply voltage of 220 V and a fuse rating of 9 A, the maximum allowable power is 1980 W (220 V × 9 A) [t2][t3]. Each bulb is rated at 60 W. Since the bulbs are connected in parallel, the total power consumed is the sum of the power of each bulb [t6]. By dividing the total power capacity (1980 W) by the power of a single bulb (60 W), we get 33 bulbs [t2][t3][t8]. Alternatively, one can calculate the current drawn by a single bulb (I = P/V), which is approximately 0.2727 A [t10]. Dividing the total fuse current (9 A) by the current per bulb (0.2727 A) also yields 33 bulbs [t2][t5]. Thus, 33 bulbs can be safely operated without blowing the fuse.