Three capacitors, C1= 6.0 uF, C2= 4.0 uF, and C3= 5.0 uF, are connected to a potential difference of 24V. Find the equivalent capacitance for the three capacitors connected in (a) Series and (b) Parallel.
To find the equivalent capacitance for capacitors connected in series, you can use the following formula:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Let's substitute the given values into the equation:
1/Ceq = 1/6.0 uF + 1/4.0 uF + 1/5.0 uF
To simplify the equation, we need to find the least common denominator (LCD) of the fractions:
LCD = 60 uF
Now, we can rewrite the equation with the LCD:
1/Ceq = (10/60 + 15/60 + 12/60) uF
Adding the fractions, we get:
1/Ceq = 37/60 uF
Invert both sides of the equation to solve for Ceq:
Ceq = 60/37 uF
Ceq ≈ 1.62 uF
So, the equivalent capacitance for the capacitors connected in series is approximately 1.62 uF.
To find the equivalent capacitance for capacitors connected in parallel, you simply add the individual capacitances:
Ceq = C1 + C2 + C3
Ceq = 6.0 uF + 4.0 uF + 5.0 uF
Ceq = 15.0 uF
Thus, the equivalent capacitance for the capacitors connected in parallel is 15.0 uF.
(a) 1/C = 1/6 + 1/4 + 1/5
(b) C = 6+4+5