Question

A 12v battery is connected in a circuit with 3 series connected resistors having resistances of 4 ohm, 9 ohm and 11 ohm. Determine the current flowing through and the voltage across the 9 ohm resistor. Also find the power dissipated in the 11 ohm resistor?

Answer 1

Explanation:

here is your answer

hope it's helpful

Answer 2

Answer:

I=1.33 A

P=13.09

Explanation:

v=IR

I=V/R

I=12/9

I=1.33 A

P=I^2R

I=V^2/R^2×R

so

P=V^2/R

P=(12)^2/11

P=144/11

P=13.09