please help:- Two capactiors of 14microfarror and 6 microfarror are connected in parallel with a series 10microfarror capacitor across a 100v dc supply

a)total capactiance
b)charge on each capacitor
c)voltage across each capacitor
d) energy stored in each capacitor

These are standard formulas: what is your qu?estion about it

the question says to evaluate and answer a,b,c,d

can i ask what are the forumulas?

To solve this problem, let's break it down step by step:

Step 1: Total Capacitance (C_total)
In parallel, the total capacitance (C_total) of capacitors is equal to the sum of their individual capacitances (C1 + C2 + ... + Cn). In this case, we have two capacitors in parallel, with individual capacitances of 14μF and 6μF.

C_total = C1 + C2
C_total = 14μF + 6μF
C_total = 20μF

So, the total capacitance of the capacitors connected in parallel is 20μF.

Step 2: Charge on each capacitor (Q)
When capacitors are connected in parallel, they share the same charge. The total charge (Q) can be calculated by multiplying the total capacitance (C_total) by the applied voltage (V).

Q = C_total * V
Q = 20μF * 100V
Q = 2000μC

The charge on each capacitor is 2000μC.

Step 3: Voltage across each capacitor (V_capacitor)
When capacitors are connected in parallel, they have the same voltage across them as the applied voltage.

V_capacitor = V
V_capacitor = 100V

The voltage across each capacitor is 100V.

Step 4: Energy stored in each capacitor (E)
The energy stored in a capacitor can be calculated using the formula:

E = (1/2) * C * V^2

For each capacitor, substitute the respective capacitance and voltage value.

For the 14μF capacitor:
E1 = (1/2) * 14μF * (100V)^2
E1 = 0.07 J

For the 6μF capacitor:
E2 = (1/2) * 6μF * (100V)^2
E2 = 0.03 J

The energy stored in the 14μF capacitor is 0.07 J, and the energy stored in the 6μF capacitor is 0.03 J.