Two capacitors of capacitance $1.0$ microfarad and $2.0$ microfarad are each charged by being corrected across a $5.0$ volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the $2.0$ microfarad capacitor?
A) $0\,V$
B) $0.6\,V$
C) $1.7\,V$
D) $3.3\,V$

Hint: Use the formula of the capacitance of the series circuit and substitute the capacitance of the two capacitors in it. Then use the formula, to find the voltage of the battery across the capacitor that is connected as a second and approximate its value.

Formula used:
The formula for the capacitance in the series is given by
${C_s} = \dfrac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}$
Where ${C_s}$ is the total resistance which is connected in series, ${C_1}$ is the first capacitance and ${C_2}$ is the second capacitance.

Complete step by step solution:
It is given that the
Capacitance of the first capacitor, ${C_1} = 1.0\,\mu F$
Capacitance of the second capacitor, ${C_2} = 2.0\,\mu F$
Potential difference of the battery, $V = 5\,V$
It is provided that the opposite charges are connected to each other. Hence the capacitor circuit must be connected in series. Hence use the formula of the capacitance in series,
${C_s} = \dfrac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}$
Substituting the capacitance of the two capacitors in it.
$\Rightarrow {C_s} = \dfrac{{1.2}}{{1 + 2}} = \dfrac{2}{3}\,\mu F$
Hence the total capacitance value is obtained as $\dfrac{2}{3}$.
The magnitude of the voltage across the $2\,\mu F$ capacitor is $\dfrac{{5\,V}}{3} = 1.667\,V$.
Hence the magnitude of the voltage across the second capacitor is $1.7V$.

Thus the option (C) is correct.

Note: In this case the current flowing through the circuit is the same in the series. But if the capacitors are connected in parallel, then the total capacitance is the sum of the value of the capacitance of the connected capacitors in the circuit.