Question

Please I need help I have an exam tomorrow and I need the answer quickly

Answer 1

(a) The equivalent capacitance of the circuit with the switch S opened is 48.67 μF.

(b) The equivalent capacitance of the circuit with the switch S closed is 140 μF.

(a) When the switch S is open, the 40μF and 70μF capacitors are in series. The equivalent capacitance
`(C_s)` of capacitors in series can be calculated using the formula:

`[ (1)/(C_s) = (1)/(C_1) + (1)/(C_2) ]`

Thus:

`\[ (1)/(C_s) = (1)/(40 \, \mu F) + (1)/(70 \, \mu F) \]`

`\[ (1)/(C_s) = (1)/(40) + (1)/(70) \]`

Taking the reciprocal:

`\[ (1)/(C_s) = (110 + 40)/(40 * 70) \]`

`\[ (1)/(C_s) = (150)/(2800) \]`

`\[ C_s = (2800)/(150) \]`

`\[ C_s = 18.67 \, \mu F \]`

This equivalent capacitance is in parallel with the 30μF capacitor. The equivalent capacitance
`(C_p)` of capacitors in parallel is the sum of the individual capacitances:

`\[ C_p = C_s + 30 \, \mu F \]`

`\[ C_p = 18.67 + 30 \]`

`\[ C_p = 48.67 \, \mu F \]`

(b) When the switch S is closed, all three capacitors are in parallel. The equivalent capacitance of capacitors in parallel is the sum of the individual capacitances:

Given:

`\[ C_p = 30 \, \mu F + 40 \, \mu F + 70 \, \mu F \]`

`\[ C_p = 30 + 40 + 70 \]`

`\[ C_p = 140 \, \mu F \]`

So, the results are:

-
`\( C_p = 48.67 \, \mu F \)` (with the switch S open)

-
`\( C_p = 140 \, \mu F \)` (when the switch S is closed)