Draw a schematic diagram of a circuit consisting of a battery of three cells of $2\, V$ each,a $5\,\Omega$ resistor,an $8\,\Omega$ resistor,and a $12\,\Omega$ resistor,and a plug key,all connected in series. Include an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the $12\,\Omega$ resistor. What would be the readings in the ammeter and the voltmeter?

(N/A) To measure the current flowing through the resistors,an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the $12\,\Omega$ resistor,a voltmeter should be connected in parallel to this resistor.
The total voltage of the battery is $V = 3 \times 2\, V = 6\, V$.
The resistances are connected in series,so the total resistance $R$ is:
$R = 5\,\Omega + 8\,\Omega + 12\,\Omega = 25\,\Omega$.
According to Ohm's law,$V = IR$,the current $I$ flowing through the circuit is:
$I = \frac{V}{R} = \frac{6\, V}{25\,\Omega} = 0.24\, A$.
Thus,the ammeter reading is $0.24\, A$.
The potential difference $V_1$ across the $12\,\Omega$ resistor is calculated using Ohm's law:
$V_1 = I \times R_{12\,\Omega} = 0.24\, A \times 12\,\Omega = 2.88\, V$.
Therefore,the reading of the voltmeter is $2.88\, V$.