Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

home / study / science / advanced physics / questions and answers / in figure 5,

ID: 1439810 • Letter: H

Question

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