Help with this EK problem?

[m25]

---

Members don't see this ad.

Why is the voltage across the capacitor 6V, not 12V? I thought capacitors were able to use the highest voltage provided by battery.
And for 158, would the current flowing through B always equal to 0, whether it's right after the switch or not, because it is part of a broken circuit?

 

[NextStepTutor_3]

The voltage across the capacitor would be 12 V if it weren't in series with that last resistor. Think about Kirchoff's loop rule - the voltage drops throughout a circuit must add up to the emf of the battery. As the capacitor is charging (and the switch is closed), the current leaving the battery is 3 A, like they say in the explanation. Let's work backwards - if the current is 3 A and the resistance of that final resistor is 2 Ohms, its voltage drop must equal (3)(2) = 6V. For the total voltage to sum to 12 V, the voltage across each parallel branch has to be 12 - 6 = 6 V.

For #158, you're completely right! When you open the switch, resistor B becomes part of a broken circuit, eliminating the current flow to that arm of the diagram. So the current through such a resistor will always be zero in cases like this.