If C = 48 mu F, determine the equivalent capacitance for the combination shown.
ID: 1649822 • Letter: I
Question
If C = 48 mu F, determine the equivalent capacitance for the combination shown. A) 20 mu F B) 36 mu F C) mu F D) 32 mu F A parallel plate capacitor of capacitance C_0 has plates of area A with separation d between them. When it is connected to a battery of voltage V_0, it has charge of magnitude Q_0 on its plates. The plates are pulled apart to a separation 2d while the battery is disconnected from the circuit. After the plates are 2d apart, the capacitance of the capacitor and the magnitude of the potential difference are: A) 1/2 C_0, V_0 B) 1/2 C_0, 2V_0 C) C_o, V_0 D) 2C_0, V_0 A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positive plate of each connected to the negative plate of the other. What is the final charge on the 4.0-mF capacitor? 0 mC B) 10 mC C) 8 mC D) 12 mCExplanation / Answer
Solution:
1.
The two pairs of capacitors of capacitance 2C are in series & then the combination is in parallel & then series with the capacitor of capacitance C:
=> C1 = {(2C)(2C) /(2C + 2C)} = C
C2 = {(2C)(2C) /(2C + 2C)} = C
C1 and C2 in parallel => C3 = C1 + C2 = C+C = 2C
=> C3 is in series with C
=> Ceq = (2C)(C)/ (2C+C)
=> Ceq = 2C/3 = 2*48/3
=> Ceq = 32uF
(Option-D)
2.
As we know;
Capacitance = A()/d
=> Capacitance is inversly proportional to d
=> C(new) = Co/2
But as the Charge does'nt change => this C(new) will affect the voltage{Q = CV}.
=> V(new) = 2V
(Option-B)