Three $2 \,\Omega$ resistors,$A, B$ and $C,$ are connected as shown in the figure. Each of them dissipates energy and can withstand a maximum power of $18 \,W$ without melting. Find the maximum current $I$ that can flow through the circuit.

(3 A) The power dissipated by a resistor is given by the formula $P = I^2 R$.
Given that the maximum power each resistor can withstand is $P_{max} = 18 \,W$ and the resistance is $R = 2 \,\Omega$.
First,calculate the maximum current $I_{max}$ that any single resistor can handle:
$I_{max}^2 = P_{max} / R = 18 / 2 = 9$
$I_{max} = \sqrt{9} = 3 \,A$.
In the given circuit,resistor $A$ is in series with the parallel combination of resistors $B$ and $C$.
The total current $I$ flows through resistor $A$. Therefore,the maximum current $I$ that can flow through the circuit is limited by the maximum current capacity of resistor $A$,which is $3 \,A$.
If $I = 3 \,A$ flows through the circuit,then at the parallel junction,the current splits equally between $B$ and $C$ (since $R_B = R_C = 2 \,\Omega$).
Thus,current through $B$ is $I_B = 1.5 \,A$ and current through $C$ is $I_C = 1.5 \,A$.
Since $1.5 \,A < 3 \,A$,both $B$ and $C$ are safe.
Therefore,the maximum current that can flow through the circuit is $3 \,A$.