A graph of charge q against voltage V is drawn for the series and the parallel combination of two given capacitors. Straight line A through the origin reaches q=500 μC at V=10 V, and straight line B through the origin reaches q=80 μC at V=10 V. The capacitances are

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(4) 40 μ F and 10 μ F. Each line has slope C=q/V: line A gives (500)/(10)=50 μ F and line B gives (80)/(10)=8 μ F. The parallel combination always stores more charge at a given voltage, so A is the parallel one: C₁+C₂=50 μ F, and B is the series one: (C₁C₂)/(C₁+C₂)=8. Hence C₁C₂=8×50=400. C₁,C₂ are the roots of x²-50x+400=0, i.e. x=(50±30)/2. The capacitances are 40 μ F and 10 μ F.

A graph of charge q against voltage V is drawn for the series and the parallel combination of two given capacitors. Straight line A through the origin reaches q=500 μC at V=10 V, and straight line B through the origin reaches q=80 μC at V=10 V. The capacitances are

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