# Three resistors A,B,C are connected as shown in the figure.Each of them dissipates energy to maximum of 18 W .Find the maximum current that can flow through three resistors.

2
by safura

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by safura

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1/Req=1/Rb+1/Rc

1/Req=1/2+1/2

Req=1Ω

This equivalent is series with A resistance

Rnet=Req+R(A)

Rnet=2+1=3Ω

As power=(Current)² Resistance

18=I²X3

I²=6 ,I=√6

I=2.2A current

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V= voltage across a resistor

I = current through the resistor

R = value of resistance.

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.

I = current through the resistor

R = value of resistance.

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.