Series and Parallels

The trick is to minimize the $I^2R$ losses of the load which are the two lamps. That is to minimize the current $I$ through the lamps. Since by Ohm's Law, $I = \dfrac{V}{R}$ to minimize $I$ , we minimize $V$ and maximize $R$ . The smallest $V$ is achieved by connecting the 9-volt batteries in parallel ( $V=9 \space V$ ). To maximize $R$ we connect the lamps in series ( $R=10\space \Omega$ ). The current is $I = 0.9 \space A$ . Therefore, the answer is $\boxed{\text{C}}$ .