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ID: 1439810 • Letter: H
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home / study / science / advanced physics / questions and answers / in figure 5, e = 15.0 kv, c = 3800 pf, r1 = r2 ...
Question
In Figure 5, E = 15.0 kV, C = 3800 pF, R1 = R2 = R3 = 6.80 M?. At time t = 0 (i.e., immediately after the circuit is connected) the current through R1 = ___________. (Give your answer in the form of "a.bc x 10^(x)" mA)
At t = 0 (i.e., immediately after the circuit is connected) the current in R2 = _______. (Give your answer in the form of "a.bc x 10^(x)" mA.)
At t = 0 (i.e., immediately after the circuit is connected) the current in R3 = _______. (Give your answer in the form of "a.bc x 10^(x)" mA.)
What is the current in R3 at t = ? ? (Give your answer in the form of "a.bc x 10^(x) A".)
What is the potential difference across R1 at t = ? ?. (Give your answer in the form of "a.bc x 10^(x) V".)
What is the potential difference across R2 a very long time after the circuit is connected? Give your answer in the form "a.bc x 10^(4)" V.
R‘ 5.Explanation / Answer
at t = 0 , no current in capacitor
Req = R1 + R3
I = e/R = 15*10^3/(6.8*2*10^6)
I = 1.1 x 10^(0) mA
since R1 and R3 in series so current is same
I1 = 1.1 x 10^(0) mA
part 2 )
at t = 0 no current in R2
I2 = 0.00 x 10^(0) mA
part c ) at R3
is equal to R1
I3 = I1 = 1.1 x 10^(0) mA
part d )
at t = infinite
here R2 and R3 in parallel
R23 = 3.4 Mohm =
R23 in series with R1
Req = 10.2 Mohm = 10.2 x 10^6 ohm
I = 1.47 x 10^-3
this total current in circuit
here R1 and R23 in series
so current will divide in R2 and R3
R3 = 1.47x10^-3/2 = 7.35 x 10^(-4) A
part e )
V = IR1
V = 1.00 x 10^(4) V
part f )
V2 = IR2
V2 = 0.50 x 10^4 V
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