Capacitors in the circuit have the capacitances C1=5.5μF, C2=4.4μF, C3=2.2μF, and C4=44μF and the emf of the battery is E=5V. a) Find the equivalent capacitance of the circuit.
Use the following constants if necessary. Coulomb constant, k=8.987×109N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10−12F/m. Magnetic Permeability of vacuum, μ0=12.566370614356×10−7H/m. The magnitude of the Charge of one electron, e=−1.60217662×10−19C. Mass of one electron, me=9.10938356×10−31kg. Unless specified otherwise, each symbol carries its usual meaning. For example, μC means microcoulomb.
PART -I
Capacitors in the circuit have the capacitances C1=5.5μF, C2=4.4μF, C3=2.2μF, and C4=44μF and the emf of the battery is E=5V.
a) Find the equivalent capacitance of the circuit.
PART-II
The capacitor in the first circuit C2 is replaced by the resistor R=11Ω in the second circuit. And a switch is placed as shown in the figure. The emf of the circuit is E=5V. Consider the capacitors are completely uncharged when the switch is open.
b) Explain briefly what will happen - if we close the switch and connect to the P terminal of the circuit. Draw the circuit diagram.
c) Calculate the time constant τ of the circuit.
Part -I
Given Data:
C1 = 5.5 uF
C2 = 4.4 uF
C3 = 2.2 uF
C4 = 44 uF
E.M.F of the battery = 5 Volt
We have to find the equivalent capacitance of the circuit
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