Question

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Answer 1

Power  dissipated by the resistor is given by : P =  V²/R = I² R

      I² = P / R

Each resistance dissipates energy = 18 Watts

Maximum current through B or C or A is given by : I² = 18 Watts/2 ohms

          So I = 3 Amperes maximum

Since the resistances B and C are connected in parallel, the current that comes from the resistance A is divided equally between the resistance B and C.

  so current through  B = current through resistance C = 1/2 * current thru A

Since, the current passing thru resistor A is max = 3 ampere, then the current through resistor B and resistor C  are 1.5 Amperes maximum.

Answer 2

Answer:

The maximum current that can flow through the three resistor is 1.5A.

Explanation:

$as \: P=VI \:$

Given P is 18W .

$ohm \: law \: V=IR \:$

P=VI

P=IR×I

P=I²R

Therefore as the resistance through B or C or A is 2 ohm.

I²=P/R

I²=18/2

I²=9

I=3A

And as B & C are in parallel connection while A in series and the current is divided so the current will also get divided as 1.5 A each through B & C.