Basic Physics II Evidences - DrJJ - UiTM

CONFIDENTIAL3AS/APR 2008/PHY407QUESTION 2a)BC 1 =6 µFAC 3 =5 µFC 2 =4 µF12 V+-CC 4 =10 µFDi) The capacitors in parallel adds up: C12 C1 C2 6 F 4 F 10 FThe capacitors in series now includes C 12 , C 3 and C 4 :1 1 1 1 1 1 1 1 2 1 4 . Then theC1234 C12C3C410F5 F10F10F10Fcapacitors can now be replaced by a single capacitor by taking the inverse:10 5C 1234 F F 2.5 F4 2ii) Since capacitance is the ratio of charge for every volt of potential across the plates,qC , then, the total charge stored in the single capacitor C 1234 isV5q CV F 12V 30 C2iii) Since the charges will flow from the battery through a single circuit then the potential dropq 30Cacross point AB must be VABV1V2 3VC 10Fthe charge stored in C 2 is:q 2122 C2V 4 F 3V 12 C(9+2+5=16 marks)b) Figure 3 shows an arrangement of resistors connected together in a circuit with a 12-voltbattery. For the following questions, leave your answers in the form of fractions. DO NOTCONVERT YOUR ANSWERS TO DECIMAL.© Hak Cipta Universiti Teknologi MARA CONFIDENTIAL