When the switch is closed, the two resistors in parallel on the right combine to give R . Whenadded in series to the resistor in the main line, this gives a new equivalent resistance for the circuit of
R . This will give a new larger total current passing through the battery and ammeter A 1 :Ammeter A 2 , however, only receives half of this total current, as the current splits evenly to pass
through each of the parallel sections on the right of the circuit:Thus, the current in A 2 decreases when the switch is closed.3 . B —When the switch is initially closed, the uncharged capacitor acts as a “wire” or “closed switch”
with no resistance. Thus, the initial circuit “looks” like a parallel circuit. As the capacitor charges,
the current through the resistor in series with the capacitor drops to zero because the capacitor acts as
a “broken wire” or “open switch” with infinite resistance. After a long time, the circuit becomes a
series circuit with current passing only through a single resistor. As the circuit transitions from this
“parallel to series,” the equivalent resistance of the circuit increases. This produces a current through
the ammeter that drops from its maximum starting value to a steady state of lower value, as shown in
choice B.
4 . B —The resistance external to the battery is 3 Ω. The internal resistance of the battery is 1 Ω. Added
together in series, the internal and external resistance is 4 Ω. This gives a total current of 6 A passing
through the battery:
The voltage drop in the external portion of the circuit will be equal to the terminal voltage:This same result can be achieved by finding the internal voltage drop inside the battery and
subtracting it from the emf of the battery: