In the circuit of the figure 8 = 3.10 kV, c = 6.70 ur, R1 = R2 = R3 = 0.730 MQ.

Question

In the circuit of the figure 8 = 3.10 kV, c = 6.70 ur, R1 = R2 = R3 = 0.730 MQ. with C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i1 in resistor 1, (b) current i2 in resistor 2, and (c) current i3 in resistor 3? At too (that is, after many time constants), what are (d)/1, (e)/2, and (f)h? what is the potential difference V2 across resistor 2 at (g)t = 0 and (h)t = oo? R2 (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number (9) Number (h) Number Units Units Units Units Units Units Units Units

Explanation / Answer

Solution :-

At time t=0,

the capacitor does not provide any resistance to the flow of current.

let i1 (throgh R1), i2(through R2) and i3 through R3

i1 = i2 + i3

and applying KVL,

-E+ i1.R1+ i2.R2=0

also, i2. R2 - i3 R3=0

solving,

a . i1 = 3.10Kv/(0.73 + 0.73/2) = 2.831mA

b. i2 = i1/2 = 2.831 mA / 2 = 1.412 mA

c. i3 = i1/2 = 2.831mA / 2 = 1.412 mA   

d. t= infinity, capacitor work as open circuit

i3 =0 , i1= i2 = 3.10Kv /(0.73*2) = 2.1233 mA

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