Question

Three capacitors each of capacitance 1 \mu F1μF are connected in parallel. To this combination, a fourth capacitor of capacitance 1 \mu F1μF is connected in series. The resultant capacitance of the system is (1) 4 \mu F4μF (2) (4 / 3) \mu F(4/3)μF (3) (4 / 3) \mu F(4/3)μF (4) 2 \mu F2μF

Answer 1

resultant capacitance of the system is 3/4  μF  

Explanation:

Three capacitors each of capacitance 1μF are connected in parallel.

Capacitors in parallel Gets added

1 + 1 + 1  = 3  μF

Now 3  μF  is in Series  with    1μF

Hence Ceq  =   1/(1/3 + 1/1)

= 3/(1 + 3)

= 3/4

resultant capacitance of the system is 3/4  μF  

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Answer 2

Explanation:

Given Three capacitors each of capacitance 1 \mu F1μF are connected in parallel. To this combination, a fourth capacitor of capacitance 1 \mu F1μF is connected in series. The resultant capacitance of the system is

• Now we need to find the resultant capacitance.
• So the three capacitors are connected in parallel and another capacitor of capacitance 1 μ F is connected in series. So it will be C equivalent. In parallel it will be added.So the three equivalent can be considered as C and another is 1 micro F.
• So C eq = C x 1 / C + 1
•            = C / C + 1  
• Therefore C = 1 + 1 + 1  
•                    = 3 μ F
• So C eq = 3 / 3 + 1
• So C eq = 3/4 μ F

Reference link will be

https://brainly.in/question/18818248