Question

Problem 26.45 Constants Periodic Table VPart A Two 4.5 μF capacitors, two 2.0 kf2 resistors, and a 10.8 V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its in Express your answer using two significant figures 10.0051 Submit Previous Answers Request Answer X Incorrect; Try Again; 7 attempts remaining Provide Feedback

Answer 1

Here, Equivalent capacitance will be:

$C_{eq} = [\frac{1}{4.5\mu}+ \frac{1}{4.5\mu}]^{-1} = 2.25\mu F$

total resistance = R = 2000 + 2000 = 4000 ohms

so, $\tau = RC_{eq} = (4000)2.25\mu =0.009 s$

After switching on the circuit, the current will drop exponentially and the trend is given by:

$I = I_oe^{-t/\tau}$

=> $\frac{I}{I_o}=e^{-t/0.009}$

now take log_e on both sides to get:

$ln\frac{I}{I_o}=-t/0.009$

=> $t = - 0.009ln\frac{I}{I_o}$

substitute I/I_o from the problem [example: if I falls by 50% of initial current then, I/I_o = 0.5] to get the time in which the current drops from maximum to the given value.