A capacitor C₁=1.0 μ F is charged up to a voltage V=60 V by connecting it to a battery B through switch (1). Now C₁ is disconnected from the battery and connected through switch (2) to a circuit of two uncharged capacitors C₂=3.0 μ F and C₃=6.0 μ F, which are in series with each other, that series…

Question

A capacitor C₁=1.0 μ F is charged up to a voltage V=60 V by connecting it to a battery B through switch (1). Now C₁ is disconnected from the battery and connected through switch (2) to a circuit of two uncharged capacitors C₂=3.0 μ F and C₃=6.0 μ F, which are in series with each other, that series pair being placed straight across C₁. The charge that finally flows into the C₂–C₃ branch is

Accepted Answer

(2) 40 μC. Initial charge: Q=C₁V=1×60=60 μC, and once the battery is disconnected this is all the charge in the circuit. C₂ and C₃ are in series: C₂₃=(3×6)/(3+6)=2 μ F. Closing switch (2) puts C₂₃ in parallel with C₁, so the total capacitance is 1+2=3 μ F and the common voltage is V'=(60)/3=20 V. Charge in the series branch: q=C₂₃V'=2×20=40 μC, and C₁ keeps 1×20=20 μC. Being in series, C₂ and C₃ each carry that same 40 μC (at 13.3 V and 6.7 V respectively).

Step-by-step solution

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