Question

A 9.00-volt battery is used to power a series 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

Answer:

we know

potential difference varies in series combination

hence we will find current first.

v=ir

v/r=I

9/6=I

I=1.5 amps

power across resistance of 2.5 ohm

p=I*l*r

=2.25*2.5

=5.625watts

power across resistance of 3.5 ohm

p=I*I*r

2.25*3.5

=7.875 watts

power of the circuit

p=I*i*r

2.25*6

=13.5watts