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The ammeter is connected in series with the battery. (a) Which terminal of the ammeter is represented by point P? (b) Calculate the ammeter reading.

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Part 3.1

Which terminal of the ammeter is represented by point P? Write down POSITIVE or NEGATIVE.

Current flows from the positive terminal of the cell, through the external circuit, and back to the negative terminal. Ammeters are connected such that current enters the positive terminal and exits the negative terminal. Point P is on the side of the ammeter closer to the negative terminal of the cell, meaning current exits the ammeter at P.
Therefore, P represents the NEGATIVE terminal.

Part 3.2· 3 steps

Calculate the ammeter reading.

  1. Calculate the equivalent resistance of the series combination of 4 Ω4 \, \Omega and 2 Ω2 \, \Omega.

    Rseries=4 Ω+2 Ω=6 ΩR_{series} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega

  2. Calculate the total external resistance (RextR_{ext}) of the circuit.

    The 6 Ω6 \, \Omega series combination is in parallel with the 3 Ω3 \, \Omega resistor.

    1Rext=16 Ω+13 Ω\frac{1}{R_{ext}} = \frac{1}{6 \, \Omega} + \frac{1}{3 \, \Omega} 1Rext=16 Ω+26 Ω=36 Ω\frac{1}{R_{ext}} = \frac{1}{6 \, \Omega} + \frac{2}{6 \, \Omega} = \frac{3}{6 \, \Omega} Rext=6 Ω3=2 ΩR_{ext} = \frac{6 \, \Omega}{3} = 2 \, \Omega

  3. Calculate the ammeter reading (ItotalI_{total}) using Ohm's law for the external circuit.

    The voltmeter reads the terminal voltage (VextV_{ext}) across the external resistance.

    Itotal=VextRext=1.36 V2 Ω=0.68 AI_{total} = \frac{V_{ext}}{R_{ext}} = \frac{1.36 \, V}{2 \, \Omega} = 0.68 \, A

    The ammeter reading is 0.68 A\boxed{\text{0.68 A}}.

Part 3.3· 2 steps

Determine the internal resistance of the cell.

  1. Use the formula for terminal voltage, Vext=ε−ItotalrV_{ext} = \varepsilon - I_{total}r.

    1.36 V=1.5 V−(0.68 A)r1.36 \, V = 1.5 \, V - (0.68 \, A)r

  2. Rearrange the equation to solve for rr.

    (0.68 A)r=1.5 V−1.36 V(0.68 \, A)r = 1.5 \, V - 1.36 \, V (0.68 A)r=0.14 V(0.68 \, A)r = 0.14 \, V r=0.14 V0.68 Ar = \frac{0.14 \, V}{0.68 \, A} r≈0.20588 Ωr \approx 0.20588 \, \Omega

    The internal resistance of the cell is 0.21 \Omega\boxed{\text{0.21 \Omega}}.

Part 3.4

An additional resistor X is connected parallel to the 3 Ω3 \, \Omega resistor in the circuit. Will the reading on the ammeter INCREASE, DECREASE or REMAIN UNCHANGED? Give a reason for the answer.

The reading on the ammeter will INCREASE.

Reason:
When an additional resistor is connected in parallel to the 3 Ω3 \, \Omega resistor, the equivalent resistance of that parallel combination decreases. This leads to a decrease in the total external resistance (RextR_{ext}) of the circuit. Since the total resistance of the circuit (Rtotal=Rext+rR_{total} = R_{ext} + r) decreases and the EMF (ε\varepsilon) remains constant, the total current (Itotal=εRtotalI_{total} = \frac{\varepsilon}{R_{total}}) flowing through the circuit (and thus the ammeter reading) must increase.

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