At what rate is thermal energy being generated in the 2R-resistor when $\varepsilon$= 12 V and R = 3.0 $\Omega$ ?
A)12 W
B)24 W
C)6.0 W
D)3.0 W
E)1.5 W

Question

At what rate is thermal energy being generated in the 2R-resistor when $\varepsilon$= 12 V and R = 3.0 $\Omega$ ? 
A)12 W 
B)24 W 
C)6.0 W 
D)3.0 W 
E)1.5 W

Quizplus · Accepted Answer

We can use the formula for power (P) in a resistor: 
P = I^2R = V^2/R
where I is the current, R is the resistance, and V is the voltage across the resistor.

First, we need to find the current through the resistor. We can use Ohm's law:
V = IR
Solving for I, we get:
I = V/R

Substituting the given values, we get:
I = 12 V / 3.0 Ω = 4.0 A

Now we can use the formula for power:
P = I^2R = (4.0 A)^2 * 3.0 Ω = 48 W

However, this is the total power in the circuit. To find the power specifically in the 2R-resistor, we need to use the voltage drop across it. We can use Kirchhoff's voltage law (KVL) to find this:
V_2R + V_R = 12 V
where V_2R is the voltage drop across the 2R-resistor and V_R is the voltage drop across the other resistor. Since they are equal in this circuit, we can simplify to:
2V_2R = 12 V
V_2R = 6.0 V

Now we can use the power formula again, with R = 2R = 6.0 Ω and V = 6.0 V:
P = V^2/R = (6.0 V)^2 / 6.0 Ω = 6.0 W

Therefore, the answer is C) 6.0 W.

At What Rate Is Thermal Energy Being Generated in the 2R-Resistor