In the given circuit, C₁=2 μ F, C₂=0.2 μ F, C₃=2 μ F, C₄=4 μ F, C₅=2 μ F and C₆=2 μ F. A 10 V cell sits in the left branch. From its positive terminal the top wire runs through C₁, then through C₃, to the top right corner; from its negative terminal the bottom wire runs through C₆, then through C₅,…

Question

In the given circuit, C₁=2 μ F, C₂=0.2 μ F, C₃=2 μ F, C₄=4 μ F, C₅=2 μ F and C₆=2 μ F. A 10 V cell sits in the left branch. From its positive terminal the top wire runs through C₁, then through C₃, to the top right corner; from its negative terminal the bottom wire runs through C₆, then through C₅, to the bottom right corner. C₂ is a vertical branch joining the C₁–C₃ junction to the C₆–C₅ junction, and C₄ is a vertical branch joining the two right-hand corners. The charge stored on capacitor C₄ is μ C.

Accepted Answer

4. Call the C₁–C₃ junction M and the C₆–C₅ junction N. Between M and N there are two paths: C₂ directly, and C₃, C₄, C₅ in series round the right-hand loop. Series branch: 1/C=1/2+1/4+1/2=5/4, so C=0.8 μ F. With C₂ in parallel, C_MN=0.8+0.2=1 μ F. The cell then sees C₁, C_MN and C₆ in series: 1/(C_eq)=1/2+1/1+1/2=2, so C_eq=0.5 μ F and the total charge is 0.5×10=5 μ C. That same 5 μ C sits across C_MN, so V_MN=5/1=5 V. The series branch carries q=0.8×5=4 μ C, and in a series branch every capacitor holds the same charge, so q_C₄=4 μ C.

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