1) A 6-V battery is connected to an RC circuit (R = 16 ? and C = 4.1 ?F). If the

Question

1) A 6-V battery is connected to an RC circuit (R = 16 ? and C = 4.1 ?F). If the capacitor is initially uncharged, what is the current in the circuit immediately after the connection is made?

2) A 10-V battery is connected to an RC circuit (R = 6 ? and C = 6 ?F). Initially, the capacitor is uncharged. What is the final charge on the capacitor (in ?C)?

3) The quantity t1/2=? ln 2 is called the half-life of an exponential decay, where ?=RC is the time constant in an RC circuit. The current in a discharging RC circuit drops by half whenever t increases by t1/2. For a circuit with R=2 k? and C=3 ?F, if the current is 6 mA at t=5 ms, at what time (in ms) will the current be 3 mA?

Explanation / Answer

1)

Immediately after the connection is made the capacitor acts as a short circuit..

Thus the circuit initially seems to consist only of the Resistor(R).

So, current , I = V/R = 6/16 = 0.375 A <-------answer

2)

After a long time, the capacitor charges fully such that the voltage across it is 10V.

So, Charge across it, Q = C*V = (6*10^-6)*10 = 6*10^-5 C = 60 uC <---------answer

3)

time constant, = R*C = 2k * 3u = (2*10^3)*(3*10^-6) = 6*10^-3 s = 6 ms

So, half life = 6ms * ln(2) = 4.2*10^-3 = 4.2 ms

So, after 4.2ms, the current will decrease to half the initial value.

So, for the current to decrease to half from(6mA to 3mA), it will take 4.2ms

So, time at which it will reach 3mA = 5ms + 4.2ms = 9.2 ms <-----------answer

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