The potential diffrence across a series combination of 4(micro)F and 12(micro)F capacitor is 240V. What is the potential diffrence across 4(micro)F capacitor?
The same charge is on both capicator.
C=qV
q=C/V
so you have 4/V4 = 12/V12
and V4+V12=240
solve for them. I get V12 is 3 times V4, or V12=180, V4=60
check all that.
To find the potential difference across the 4μF capacitor in a series combination, you can use the concept of capacitors in series.
In a series combination of capacitors, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of individual capacitances. So, in this case:
1 / C_total = 1 / C1 + 1 / C2
Given C1 = 4μF (microfarads) and C2 = 12μF, we can substitute these values into the formula:
1 / C_total = 1 / 4μF + 1 / 12μF
To get the total capacitance, you need to calculate the reciprocal of both sides:
C_total = 1 / (1 / 4μF + 1 / 12μF)
Now, we can find the potential difference across the 4μF capacitor. Since the potential difference across each component in a series combination is the same, the potential difference across the 4μF capacitor is equal to the total potential difference.
Given the total potential difference (V_total) is 240V, the potential difference across the 4μF capacitor (V1) is also 240V.
Therefore, the potential difference across the 4μF capacitor is 240V.