Question

A 9.00-volt battery is used to power a parallel circuit with a 2.50 Ω and a 3.50 Ω resistor. Determine the power rating of each resistor and the total power of the circuit.

Answer 1

power =9 w

R1=2.50ohm

R2=3.50ohm

1/Req=1/R1+1/RE

1/Req=1/2.5+1/3.5

1/Req=10/25+10/35

1/Req=(70 +50)/175

1/Req=120/175

Req=175/120

Req=1.458ohm

So, equivalent resistance=1.458ohm

now

ATQ,

V=I1 ❌ R1

so, I1 =V/R1

I1=9/2.5

I1=3.6Amp

hence , P1 =I1^2 ❌ R1

P1=3.6^2❌2.5

P1=12.96watt

now,

V=I2❌R2

so, I2=V/R2

I2=9/3.5

I2=2.57Amp

hence, P2=I2^2❌R2

P2=6.60❌3.5

P2=23.1watt

In the same way P equivalent using R equivalent.

#avi taking help from pritijain.

Answer 2

Answer:

Power of 2.5 Ω =32.4 W

Power of 3.50 Ω  = 23.1 W

Total Power = 55.5 W

Explanation: