A30 resistor is connected in parallel with a 6.0 q resistor. this combination is connected in series with a 4.00 resistor. the resistons are connected to a 12. volt battery. how much power is dissipated in the 3.0 resistor? a) 7.7 w e) 5.3 w do 12 w b) 2.7 w c) 6 w

Power dissipated in 3 ohms resistor is 5.32 watts                

Explanation:

Resistor 1, R₁ = 3 ohms

Resistor 2, R₂ = 6 ohms

Resistor 3, R₃ = 4 ohms

Voltage source, V = 12 V

We need to find the power dissipated in the 3 ohms resistor. Firstly, we will find the equivalent resistance of R₁ and R₂.

R' = 2 ohms

Now R' is connected in series with R₃. Their equivalent is given by :

Total current flowing through the circuit,

Voltage across R',

The voltage across R₁ and R₂ is 4 volts as they are connected in parallel. So, current across 3 ohm resistor is,

Power dissipated is given by, P = I × V

P = 5.32 watts

So, 5.32 watt of power is dissipated  in 3 ohms resistor. Hence, this is the required solution.

battery with hi vtge high val resist in series