## SAT Physics Subject Test

**Chapter 10 ****Direct Current Circuits**

**RESISTANCE–CAPACITANCE (RC) CIRCUITS**

Capacitors are typically charged by batteries. Once the switch in the diagram on the left is closed, electrons are attracted to the positive terminal of the battery and leave the top plate of the capacitor. Electrons also accumulate on the bottom plate of the capacitor, and this continues until the voltage across the capacitor plates matches the emf of the battery. When this condition is reached, the current stops and the capacitor is fully charged.

12. Find the charge stored and the voltage across each capacitor in the following circuit, given that ε = 180 V, C_{1} = 30 μ F, *C*_{2} = 60 μ F, and *C*_{3} = 90 μ F.

Here’s How to Crack It

Once the charging currents stop, the voltage across *C*_{3} is equal to the voltage across the battery, so *V*_{3} = 180 V. This gives us *Q*_{3} = *C*_{3}*V*_{3} = (90μF)(180 V) = 16.2 mC. Since *C*_{1} and *C*_{2} are in series, they must store identical amounts of charge, and, from the diagram, the sum of their voltages must equal the voltage of the battery. So if we let *Q* be the charge on each of these two capacitors, then *Q* = *C*_{1}*V*_{1} = *C*_{2}*V*_{2} and *V*_{1} + *V*_{2} = 180 V. The equation *C*_{1}*V*_{1} = *C*_{2}*V*_{2} becomes (30μF)*V*_{1} = (60μF)*V*_{2}, so *V*_{1} = 2*V*_{2}. Substituting this into *V*_{1} + *V*_{2} = 180 V gives us *V*_{1} = 120 V and *V*_{2} = 60 V.The charge stored on each of these capacitors is

(30μF)(120 V) = *C*_{1}*V*_{1} = *C*_{2}*V*_{2} = (60μF)(60 V) = 3.6 mC

13. In the diagram below, *C*_{1} = 2 mF and *C*_{2} = 4 mF. When switch S is open, a battery (which is not shown) is connected between points *a* and *b* and charges capacitor *C*_{1} so that *V** _{ab}* = 12 V. The battery is then disconnected.

After the switch is closed, what will be the common voltage across each of the parallel capacitors (once electrostatic conditions are reestablished)?

Here’s How to Crack It

When *C*_{1} is fully charged, the charge on (each of the plates of) *C*_{1} has the magnitude *Q* = *C*_{1}V = (2 mF)(12 V) = 24 mC. After the switch is closed, this charge will be redistributed in such a way that the resulting voltages across the two capacitors, *V*′, are equal. This happens because the capacitors are in parallel. So if *Q*′_{1} is the new charge magnitude on *C*_{1} and *Q*′_{2} is the new charge magnitude on *C*_{2}, we have *Q*′_{1} + *Q*′_{2} = *Q*, so *C*_{1}*V*′ + *C*_{2}*V*′ = *Q*, which gives us