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

Answer:

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.

$\dfrac{1}{R'}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\dfrac{1}{R'}=\dfrac{1}{3}+\dfrac{1}{6}$

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, $I=\dfrac{12}{6}=2\ A$

Voltage across R', $V'=IR'=2\times 2=4\ V$

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

$I=\dfrac{4}{3}=1.33\ A$

Power dissipated is given by, P = I × V

$P=1.33\ A\times 4\ \Omega$

P = 5.32 watts

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