for the dircul shown t e \'with S is egen and he capacitor-urdwged The switch S

Question

for the dircul shown t e 'with S is egen and he capacitor-urdwged The switch S is ck sed attmet-anthe igure, when the time t is equal to 8.0 s, the charge on the capacitor, in jC, is closest to ANSWER 490 310 250 370 430 Multiple Choice Question 19.3 Part A How much current will be flowing through a 44.0 Vrm m) length of copper wire with radius 7.0 Vfrm mm) if it is conected to a source spplying 11.oormvv me resistivity ofcopper isd1.68 umes 1016)Omegaladot vm mln ANSWER 2300 Vrm Aj) 1400 fm A)) 1100 Vfrm Aj) 4600 V(frm Ap) Multiple Choice Question 19.13 Part A A 9 1(m V) battery is connected across two resistors in series. If the resistors have resistances of 660 fIrm VOmega) and 350 (fm Omega)), what is the voltage drop across the 350 Vrm 10mega) resistor? ANSWER D-6149302 4/5

Explanation / Answer

Given

R C circuit

with v = 30 V , C = 50*10^-6 F , R = 0.7*10^6 ohm

t = 8.0 s

the total charge in the circuit at t=0 s is Q = C*V

Q = 50*10^-6*30 = 1500*10^-6 C

in charging condition of a capacitor in RC circuit  

Q(t) = Q (1-e^(t/RC))

time constant T = R*C = 0.7*10^6*50*10^-6 s = 35 s

Q(8) = 1500*10^-6 (1-e^(-8/35)) C  

Q(8) = 0.0003064958075 C

Q(8) = 306.49580 *10^-6 C = 310 micro coulombs

answer is 310 micro coulombs

19.3

l = 44 m , r = 7 mm

v = 11 V , resitivity is rho = 1.68*10^-8 ohm m

current throug the wire is I = V/R

where R = rho*l/A

R = rho*l/(pi*r^2)

R = ( 1.68*10^-8*44)/(pi*(7*10^-3)^2) ohm

R = 0.004802 ohm

current is I = V/R

I = 11/(0.004802) A

I = 2290.7122 A

I = 2300 A

19.13

V = 9 V

R1 = 660 ohm

R2 = 350 ohm

R1,R2 are in seires so the net resistance is R = R1+R2 = 660+350 ohm = 1010 ohm

V = I*R

I = V/R

I = 9/1010 A

I = 0.0089108910891 A

now the potential across the resistor R2 is  

V2 = i*R2 = 0.0089108910891*350 V = 3.12 V

and V1 = I*R1 = 0.0089108910891*660 V = 5.88 V

so V = V1+V2 = 9 V

so the potential difference across the resistor 350 ohm is = 3.12 V

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