Question

1/2 means that the voltage (or charge) of the system will increase to half more of what is left in a time equal to t1/2 seconds. Therefore if a system is already at half charge (t1/2 seconds after starting) then after t1/2 more seconds the system will be charged to 50% plus half of 50%. That is 25% more, or 75% of the entire charge. Let's say that four t1/2's have gone by. That means that the charge (or voltage) is at (50% + 1/2*50% + 1/2*1/2*50% + 1/2*1/2*1/2*50%) = 93.75% of maximum charge. Yikes! Now look in your manual for a more simple mathematical derivation of this concept. Given t1/2 to be 0.4406 seconds, how long should it take to reach 75% of maximum charge? answer in seconds.

Answer 1

The answer is "0.047".

Step-by-step explanation:

Given value:

`\to t_{(1)/(2)}= 0.4406\\\\\to V=0.9375 V_(max)`

Calculating the capacitance:

`\to V=V_(max) (1-e^{(-t)/(Rc)})`

In this, the t = time, which is taken to calculates its maximum voltage.

`\to 0.9375 V_(max) = V_(max)(1-e^{- (t)/(103*(165.279*10^(-6)))})\\\\\to e^{- (t)/(103*(165.279*10^(-6)))}= 0.0625\\\\\to - (t)/(103*(165.279*10^(-6))) = \ln(0.0625)\\\\\to -t= 0.01702*(-2.77258) \\\\ \to -t = -0.04719 \\\\ \to t= 0.04719 \approx 0.047 \ s`