ln the circuit shown in figure the potential difference across the 4.5 μF capacitor is

Let us calculate the total capacitance of the circuit. The capacitors of capacitances 3 μF and 6 μF arc in parallel, Their capacitance C' is given by C′=3μF+6μF=9μFThe capacitor of capacitance C' is in series with the capacitor of capacitance 4.5 μF. Therefore, the capacitance of the system is given by 1C=19+14⋅5=13or  C=3μFThe charge flowing through the circuit=3×12=36μC∴Potential difference a cross 4.5 μF capacitor=qC=364⋅5=8 volt