Three capacitors with C1=C2=C3= 10.00 uF are connected to battery with voltage V= 194 V. What is the voltage across capacitor C1
If the capacitors are connected in series:
V1 = V2 = V3 = 194/3 = 64.67 Volts.
If they are connected in parallel:
V1 = V2 = V3 = 194 Vols.
The above answer has a calculation mistake..it should be 6.467 when in series. the answer for capacitors connected in parallel is correct.
To find the voltage across capacitor C1 in this circuit, we can use the concept of series capacitors.
When capacitors are connected in series, the total capacitance (Cs) is inversely proportional to the sum of the inverses of their individual capacitances. Mathematically, it can be expressed as:
1/Cs = 1/C1 + 1/C2 + 1/C3
In this case, as C1, C2, and C3 are equal, we can simplify the equation to:
1/Cs = 1/C1 + 1/C1 + 1/C1
Now substituting the values:
1/Cs = 1/10.00 uF + 1/10.00 uF + 1/10.00 uF
Simplifying further, we get:
1/Cs = 3/10.00 uF
To find Cs, we invert the obtained value:
Cs = 10.00 uF / 3
Now that we have the total capacitance, we can calculate the voltage across C1 by using the formula for capacitors in series connected to a battery:
V1 = (C1 / Cs) * V
Substituting the values:
V1 = (10.00 uF / (10.00 uF / 3)) * 194 V
Simplifying this expression, we finally get:
V1 = 3 * 194 V
V1 = 582 V
Therefore, the voltage across capacitor C1 is 582 V.