A circuit contains two resistors (10 Ohm & 20 Ohm) and two capacitors (12 microf

Question

A circuit contains two resistors (10 Ohm & 20 Ohm) and two capacitors (12 microfarads & 6 microfarads) connected to a 6 V battery, as shown in the diagram above. The circuit has been connected for a long time.

a) calculate the total capacitance of the circuit

b) calculate the current in the 10 Ohm resistor

c) calculate potential difference between points A and B

d) calculate the charge stored on one plate of the 6 microfarads capacitor

e) the wire is cut at point P. Will the potential difference betwwen points A and B increase, decrease, or stay the same? explain

Explanation / Answer

Given Data : Resistances R1 = 10    R2 = 20 Capacitances : C1 = 12 F = 12*10^-6 F C2 = 6 F = 6*10^-6 F Battery emf E = 6 V (a) Capacitors are in series, so the total capacitance of the circuit                     C = (C1C2)/(C1 + C2)                         = (12 F * 6 F)/(12 F + 6 F)                         = 4 F                         = 4*10^-6 F (b) Total resistance of the circuit R = R1 + R2 = 10 + 20 = 30 Current in the 10 resistor, I = E/R                                                 = 6 V/30                                                 = 0.2 A (c) Potential drop between the points A and B, V = E - IR1                                                                          = 6 V - (0.2 A * 10 )                                                                          = 6 V - 2 V                                                                          = 4 V (d) Charge on 6 F capacitor, q = CV = ( 4 F )(4 V) = 16 F (e) If you cut the wire at point p then also potential difference between A and B will be same, because capacitors are in parallel with the resistor
     

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