1) Two 3.3- ? F capacitors, two 2.5k? resistors, and a 18.0-V source are connect
ID: 1321626 • Letter: 1
Question
1) Two 3.3-?F capacitors, two 2.5k? resistors, and a 18.0-V source are connected in series.Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.00mA ?
2)An electronic circuit has two 12-k? resistors, R1 and R2, connected in series as shown in (Figure 1) . The battery voltage is 10V and it has negligible internal resistance. A voltmeter is connected across R1, as shown in (Figure 2) .
What internal resistance should the voltmeter have to be in error by less than 4{ m %} ?
1) Two 3.3-?F capacitors, two 2.5k? resistors, and a 18.0-V source are connected in series.Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.00mA ? 2)An electronic circuit has two 12-k? resistors, R1 and R2, connected in series as shown in (Figure 1) . The battery voltage is 10V and it has negligible internal resistance. A voltmeter is connected across R1, as shown in (Figure 2) . What internal resistance should the voltmeter have to be in error by less than 4{rm %} ?Explanation / Answer
1) 1/Cnet = 1/C + 1/C
Cnet = C/2 = 3.3/2 x 10-6 = 1.65 x 10-6 F
Rnet = R + R = 2R = 2 x 2.5k = 5k = 5000 Ohm
time constant = RC = 1.65 x 10-6 x 5000 = 8.25 x 10-3 = 8.25 mSec
for charging :
i = i0 e^ ( -t/RC)
1x 10-3 = (18 / 5000) e^ (-t / 8.25mSec)
e^ (-t / 8.25mSec) = 0.278
-t / 8.25 mS = -1.28
t = 10.56 m Sec or 10.56 x 10-3 Sec
2) voltage across R1 = iR1
and i = V / ( R1 + R2) = 0.416 mA
VR1 = 0.416 mA x 12k = 5 volt
if we connect a voltmeter with internal resistance r then
current flow in r .
Ir = i * R1 / (r + R1) = 0.416 x 12 / (r + 12k)
Ir = 5 / (r + 12k)
Voltage reading = r * Ir = 5r / (r + 12k)
error = 5 - 5r / (r + 12k) = 5 x 4 /100
5r / (r + 12k) = 4.8
r = 288 k Ohm or 288 x 10^3 Ohm
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