Two- capacitors A and B are connected in series across a 100 V supply and it is observed that thepotential difference across them are 60 V and 40 V- A capacitor of 2- F capacitance is now connected in parallel with A and the potential difference across B rises to 90 V- Determine the capacitance of A and B-

In series, potential difference distributes in inverse ratio of capacitance. ∴V_AV_B=C_BC_A=C_2C_1=6040=32 ∴ C_2=1.5 C_1 .....(i)Now V'_AV'_B=C'_ ...

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Byju's Answer

Standard XII

Physics

Capacitance of Parallel Plate Capacitor

Two, capacito...

Question

Two, capacitors A and B are connected in series across a 100 V supply and it is observed that the
potential difference across them are 60 V and 40 V. A capacitor of $2 μ F$ capacitance is now connected in parallel with A and the potential difference across B rises to 90 V. Determine the capacitance of A and B.

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Solution

In series, potential difference distributes in inverse ratio of capacitance.

$∴ \frac{V_{A}}{V_{B}} = \frac{C_{B}}{C_{A}} = \frac{C_{2}}{C_{1}} = \frac{60}{40} = \frac{3}{2}$

$∴ C_{2} = 1.5 C_{1}$ .....(i)

Now $\frac{V_{A}^{'}}{V_{B}^{'}} = \frac{C_{B}^{'}}{C_{A}^{'}}$

or $\frac{10}{90} = \frac{C_{2}}{(C_{1} + 2)}$

or $C_{1} + 2 = 9 C_{2}$ ....(ii)

Solving Eqs. (i) and (ii), we get

$C_{1} = 0.16 μ F$ and $C_{2} = 0.24 μ F$

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In series, potential difference distributes in inverse ratio of capacitance.∴VAVB=CBCA=C2C1=6040=32∴C2=1.5C1 .....(i)Now V′AV′B=C′BC′Aor 1090=C2(C1+2)or C1+2=9C2 ....(ii)Solving Eqs. (i) and (ii), we getC1=0.16μF and C2=0.24μF