Question

A 30 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

Answer 1

Power dissipated in 3 ohms resistor is 5.32 watts

Step-by-step 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.

`(1)/(R')=(1)/(R_1)+(1)/(R_2)`

`(1)/(R')=(1)/(3)+(1)/(6)`

R' = 2 ohms

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

`R_(eq)=R'+R_3`

`R_(eq)=2+4`

`R_(eq)=6\ ohms`

Total current flowing through the circuit,
`I=(12)/(6)=2\ A`

Voltage across R',
`V'=IR'=2* 2=4\ V`

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

`I=(4)/(3)=1.33\ A`

Power dissipated is given by, P = I × V

`P=1.33\ A* 4\ \Omega`

P = 5.32 watts

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