In the circuit of Fig: C = 6.5uf R1 = R2 = R2 = 0.73M? When C is fully discharge
ID: 2244339 • Letter: I
Question
In the circuit of Fig:
C = 6.5uf
R1 = R2 = R2 = 0.73M?
When C is fully discharged, the switch S is closed at t = 0
a) For t = 0, which is i1 on the resistor R1.
b) The current i2 in the resistor R2.
c) The current i3 in resistor R3.
d) For t = ?, which is the value of i1, i2 and i3.
Explanation / Answer
at t=0,
the capacitor does not provide any resistance to the flow of current.
hence we apply KVL for the two loops,
let i1=sa, i2=b and i3=c
so applying KCL,
a=b+c
and applying KVL,
-E+Ra+Rb=0
also,
-E+Ra+Rc=0
solving, we get i1=2E/(3R)
i2=E/(3R)
i3=E/(3R)
d) at t=infinity, the capacitor provides infinite resistance to the flow of current hence,
i3=0
i1=i2=E/(2R)