A circuit has a section AB shown in figure. The emf of the cell is 10 V and the capacitors have capacitances C1=1μF and C2=2μF. The potential difference VAB=5V. Find the charges on the capacitors (in μC).

Let the charge on the left plate of the capacitor is q, hence the charge on right plate of capacitor C2 will be -q as charge supplied by negative plate of capacitor will be -q. Moving from point A in the section of circuit AB VA−q1+10−q2=VBVA−VB=q+q2−10⇒5=32q−10Hence the charges on the capacitors q=10μC