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Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

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Resisters are connected in series.
So, the net resistance in the circuit = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Net potential = 6 V
Using Ohm’s law V = IR, we have
6 = I × 25 ⟹= 6/25=0.24
Now for the 12 Ω resistor, current = 0.24 A
So, using Ohm’s law V = 0.24 × 12 V = 2.88 V
Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.

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