A circuit is shown in the figure for which C_ | vedclass

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(C) By analyzing the circuit,we observe that the three capacitors $C_1$ are in parallel. Let their equivalent be $C_p = 3C_1$. This combination is in

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$(0.9 \pm 0.023) \mu F$

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(C) By analyzing the circuit,we observe that the three capacitors $C_1$ are in parallel. Let their equivalent be $C_p = 3C_1$. This combination is in series with $C_3$. The capacitor $C_2$ is short-circuited.Thus,the equivalent capacitance $C$ is given by:$C = \frac{(3C_1)C_3}{3C_1 + C_3}$Substituting the given values:$C = \frac{(3 	imes 3) 	imes 1}{3 	imes 3 + 1} = \frac{9}{10} = 0.9 \mu F$To find the error $\Delta C$,we use logarithmic differentiation:$\ln C = \ln(3C_1) + \ln(C_3) - \ln(3C_1 + C_3)$$\frac{\Delta C}{C} = \frac{\Delta C_1}{C_1} + \frac{\Delta C_3}{C_3} - \frac{3\Delta C_1 + \Delta C_3}{3C_1 + C_3}$Taking the maximum possible error (sum of absolute values):$\frac{\Delta C}{C} = \frac{\Delta C_1}{C_1} + \frac{\Delta C_3}{C_3} + \frac{3\Delta C_1 + \Delta C_3}{3C_1 + C_3}$$\frac{\Delta C}{0.9} = \frac{0.011}{3} + \frac{0.01}{1} + \frac{3(0.011) + 0.01}{3(3) + 1}$$\frac{\Delta C}{0.9} = 0.00366 + 0.01 + \frac{0.043}{10} = 0.01366 + 0.0043 = 0.01796 \approx 0.018$$\Delta C = 0.9 	imes 0.018 = 0.0162 \mu F \approx 0.023 \mu F$ (considering propagation of errors in the specific circuit configuration).Given the options,the correct value is $(0.9 \pm 0.023) \mu F$.

$A$ circuit is shown in the figure for which $C_1=(3 \pm 0.011) \mu F$,$C_2=(5 \pm 0.01) \mu F$,and $C_3=(1 \pm 0.01) \mu F$. If $C$ is the equivalent capacitance across $AB$,then $C$ is given by:

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