Question

A 6.0-ω and a 12-ω resistor are connected in parallel across an ideal 36-v battery. What power is dissipated by the 6.0-ω resistor?

Answer 1

The power dissipated by the 6.0-ω resistor is 6 watts.

Step-by-step explanation:

In a parallel circuit, the voltage across each resistor is the same. So, the power dissipated by the 6.0-ω resistor can be calculated using the equation P = V²/R, where P is the power, V is the voltage, and R is the resistance. In this case, the voltage is 36 V and the resistance is 6.0-ω (which is equivalent to 6 ohms). Plugging in these values, we get:

P = (36 V)² / 6 ω = 6 W

Therefore, the power dissipated by the 6.0-ω resistor is 6 watts.

Answer 2

P = 216 Watts

Step-by-step explanation:

Given that,

Resistance 1,
`R_1=6\ \Omega`

Resistance 2,
`R_2=12\ \Omega`

Voltage, V = 36 V

We need to find the power dissipated by the 6 ohms resistor. In case of parallel circuit, voltage throughout the circuit is same while the current is different.

Firstly, lets find current of the 6 ohms resistor such that,

V = IR

`I=(V)/(R)\\\\I=(36)/(6)\\\\I=6\ A`

The power dissipated by 6 ohms resistor is given by :

`P=I^2R`

`P=6^2* 6\\\\P=216\ W`

So, the power dissipated across 6 ohm resistor is 126 watts.