Upgrade to access all content for this subject

In the circuit above, $R_1 = 2R_3$ and $R_3 = 2R_2$. If the source voltage is $V_1$, find the power dissipated in $R_1$.

$$P=\frac{36V_1^2}{196R_2}$$

$$P=\frac{V_1^2}{R_1}$$

$$P=\frac{3V_1^2}{14R_2}$$

$$P=\frac{6V_1^2}{14R_2}$$

$$P=\frac{3V_1^2}{196R_2}$$