After you stabilized Johnny 5, team leader Melissa Lewis decides Artemis 13 needs to go home before anything else happens. Just as she gives the order to rotate boosters in order to send the ship home, the crew feels the spacecraft rock violently and warning lights start going off everywhere. The command ship starts to lose all of its power (maybe it had something to do with not being able to get the special tape from the STICKYSTUFF corporation at Lakeview). After a quick investigation, it turns out that the circuit that controls all of the key systems is currently drawing too much current from the battery. You are in constant communication with Austin about the issues you are encountering and a bright but intense intern proposes a risky but possibly brilliant solution. They want to take two batteries from the mostly dead Command Ship and use them to help power the Artemis. In addition to the 230 V main battery, they say they need 140 V ($V_{bat1}$) from the Command Ship battery and only 5 V ($V_{bat2}$) from the backup battery to get them home. From the ship manual you find the following resistances for different components that need to be powered in order to get you all home:

• Gauge Lights: $R_{GL}=100\Omega$
• Navigation System: $R_{NS}=700\Omega$
• Steering/Direction Control: $R_{SDC}=950\Omega$
• Engine Cooling System: $R_{ECS}=500\Omega$
• Air Filter: $R_{AF}=300\Omega$
• Waste Management System: $R_{WMS}=425\Omega$

Will this circuit work without drawing any more than the 0.35 A from the main battery?

The other suggestion on the table is rather than putting in two back up batteries, you replace them with capacitors. What would be the effect on the circuit if this were the case? Would it be better to use batteries or capacitors to get the astronauts home?