Draw a circuit diagram for a circuit consisting of a battery of five cells of $2 \text{ V}$ each, a $5 \, \Omega$ resistor, a $10 \, \Omega$ resistor, and a $15 \, \Omega$ resistor, an ammeter, and a plug key; all connected in series. Also, connect a voltmeter to record the potential difference across the $15 \, \Omega$ resistor and calculate:
$(i)$ the electric current passing through the above circuit and
$(ii)$ the potential difference across the $5 \, \Omega$ resistor when the key is closed.

(A) The circuit diagram is as shown in the provided image.
Given:
Total voltage $V = 5 \times 2 \text{ V} = 10 \text{ V}$.
Resistors in series: $R_{1} = 5 \, \Omega$, $R_{2} = 10 \, \Omega$, $R_{3} = 15 \, \Omega$.
The equivalent resistance $R$ of the series combination is:
$R = R_{1} + R_{2} + R_{3} = 5 + 10 + 15 = 30 \, \Omega$.
$(i)$ The electric current $I$ passing through the circuit is:
$I = V / R = 10 \text{ V} / 30 \, \Omega = 1/3 \text{ A} \approx 0.33 \text{ A}$.
$(ii)$ The potential difference $V_{1}$ across the $5 \, \Omega$ resistor is:
$V_{1} = I \times R_{1} = (1/3) \text{ A} \times 5 \, \Omega = 5/3 \text{ V} \approx 1.67 \text{ V}$.