Answer :

We know that,

Voltage dropor potential difference) across capacitor is given by

Where,

Q=charge on the capacitor

C=capacitance of the capacitor

By looking at the graph,

We can see that first increment in voltage is greater than the second increment. Therefore,

we can conclude that voltage drop across capacitor C_{1} is greater than the voltage drop across capacitor C_{2}

on moving left to right C_{1} comes first)

Since charges on the capacitors in series are same ,

∴ Q_{1}=Q_{2}

The potential drop across the capacitor C_{1} is more than Capacitor C_{2}.

V_{1}>V_{2}

Putting the values of V, we get

On solving, we get

Or

C_{2}>C_{1}

Thus, the capacitance of the capacitor C_{1} is less than C_{2}

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