Three 2,Omega resistors,A,B,and C,are connect | vedclass

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(A) The maximum power $P_{max}$ that each resistor can withstand is $18\,W$,and the resistance $R$ of each is $2\,\Omega$.Using the formula $P = I^2R$

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(A) The maximum power $P_{max}$ that each resistor can withstand is $18\,W$,and the resistance $R$ of each is $2\,\Omega$.Using the formula $P = I^2R$,the maximum current $I_{max}$ that any individual resistor can withstand is:$I_{max} = \sqrt{\frac{P_{max}}{R}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3\,A$.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 $I = 3\,A$,the current through $B$ and $C$ (which are equal) will be $I_B = I_C = \frac{I}{2} = 1.5\,A$.Since $1.5\,A Thus,the maximum current that can flow through the circuit is $3\,A$.

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 (in $A$) that can flow through the circuit such that none of the resistors melt.

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