A 4.19-F and a 7.46-F capacitor are connected in series across a 35.0-V battery. A 8.46-F capacitor is then connected in parallel across the 4.19-F capacitor. Determine the voltage across the 8.46-F capacitor.
To determine the voltage across the 8.46-F capacitor, we need to analyze the circuit step by step.
First, let's calculate the total capacitance (C_total) of the capacitors connected in series. When capacitors are connected in series, the reciprocal of the total capacitance is equal to the sum of the reciprocals of each individual capacitance.
Using the formula:
1/C_total = 1/C1 + 1/C2
where C1 = 4.19 F and C2 = 7.46 F
1/C_total = 1/4.19 + 1/7.46
Calculating the sum:
1/C_total = 0.2382 + 0.1341
1/C_total = 0.3723
C_total = 1/0.3723
C_total = 2.687 F (approximately)
Now, we have a single equivalent capacitor (C_total = 2.687 F) connected in series with the 8.46-F capacitor. Capacitors connected in series have the same voltage across them.
Since the total voltage of the series circuit is given as 35.0 V, the voltage across the 8.46-F capacitor is also 35.0 V.
Therefore, the voltage across the 8.46-F capacitor is 35.0 V.
so now you have 12.64-F in series with a 7.46.
Because the two equivalent series C have the same charge, voltage divides then to give across the indicated capacitor branch as
C=Q/V or VC=Q ,or v is inversely proportional to C because both branches have the same q.
V=35*12,64/(12.64+7.46)