Three `2 Omega` resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can with stand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors ?

Explanation

Resistance, `R = 2 Omega`
Maximum power, `P_(max) = 18 W`
Maximum current, `I_(max) =` ?
`P = I^(2)R`
`rArr " " I= sqrt((P)/(R)) = sqrt((18)/(2)) = 3A= I_(max)`
Maximum current that can flow through `2 Omega` resistor is `3A`. This current divides along B and C because in parallel combination, voltage across B and C remain same and hence, `I prop (1)/(R)`.
Since, B and C have same resistance `2 Omega` each, therefore, same current i.e., `(3)/(2) = 1.5 A` flows through B and C.