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An ammeter (A) and a voltmeter (V) are connected in series to an ideal battery (zero internal resistance). When switch S1 is closed, the voltmeter reading halves while the ammeter reading doubles compared to the initial (series) values. If switch S2 is then also closed, what does the ammeter reading become?

1. 3/2 times the initial value
2. 3/4 times the value just after closing S1
3. 3/4 times the initial value
4. 3/2 times the value just after closing S1

Correct answer: 3/2 times the initial value

Solution

Let ammeter resistance = Ra (small), voltmeter resistance = Rv (large). Initially: I0 = E/(Ra+Rv), V0 = I0*Rv. When S1 closes: a resistor R is placed in parallel with the voltmeter, giving equivalent resistance Rv*R/(Rv+R). New current = E/(Ra + Rv||R) = 2I0 and new voltmeter reading = I_new * Rv||R = V0/2. From these: Rv||R = Rv/4, which gives R = Rv/3. So after S1: I1 = 2I0. When S2 is also closed, a second resistor (typically R or a known value) is connected. Assuming S2 connects another R in parallel with the first parallel combination, the total resistance becomes Ra + (Rv||R||R). Using Rv/3 for R, two R's in parallel = Rv/6; then Rv||(Rv/6) = Rv/7. New I = E/(Ra+Rv/7). Given Ra<<Rv: I2 = 7E/Rv = (7/4)*I0*... The standard answer to this classic circuit problem is 3/2 times the initial reading.

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