In the given circuit calculate the power consumed in watts in the resistor of 2 Ω:

From the figure,

• $\text R_1$ = 1 Ω
• $\text R_2$ = 2 Ω
• $\text V$ = 6 V

As, $\text R$1 and $\text R2$ are connected in series then their net resistance is given by,

$\text R$\text S = \text R1 + \text R2 \\[1em] = 1 + 2 \\[1em] \Rightarrow \text R\text S = 3\ \text Ω

By using Ohm's law the current drawn from the battery is given by,

$\text I = \dfrac{\text V}{\text R_\text S} \\[1em] = \dfrac{6}{3} \\[1em] \Rightarrow \text I = 2\ \text A$

Now, power consumed by 2 Ω is given by,

$\text P = \text I^2 \text R_2 \\[1em] = 2^2 \times 2 \\[1em] = 4\times 2 \\[1em] \Rightarrow \text P = 8\ \text W$

Hence, the power consumed in the resistor of 2 Ω is 8 W.