In the above circuit, I am to calculate the max power that can be dissipated in RL. By killing the sources, the thevenin resistance is 2 kΩ. With an open at RL, the current flowing through the 2 kΩ resistor is 3 mA * ((4 / (4 + 2 + 3)) kΩ) which is 4/3 mA. Since we know the node to the left of the open is 6 V, the thevenin voltage will be 6 V minus whatever the voltage to the right of the open is. That voltage will be 4/3 mA * 3kΩ so that's 4 V. Hence, the thevenin voltage is 2 V. So shouldn't max power be Vth^2 / Rth = (2 V)^2 / 2 kΩ ??? That's 2 mW but the solution I read says 0.5 mW. The solutions I'm reading don't show the steps clearly but it says that in the above formula, Rth should be multiplied by 4 in the denominator. Where would that come from?