Can you handle this

Electricity and Magnetism Level 2

A capacitor is charged to a potential of $10V$ and it is then connected to a voltmeter having internal resistance $1M \Omega$ . After 5 seconds the voltmeter reads $5V$ , What is it's capacitance?

Note- if your answer is $x×10^{-6}$ submit value of $x$ .

none of these choices

3.5

1.5

1.99

1.7

1.4

2.3

1.9

3 solutions

Akhil Bansal
Dec 11, 2015

During discharging of capacitor,
$\large V = V_o e^{\frac{-t}{CR}}$ $\large 5 = 10 e^{\frac{-5}{10^6C}}$ Taking log both sides and simplifying, $\large C = 7.21 \times 10^{-6}$

Don't know why none of them was set as an answer to this question.

Lu Chee Ket - 5 years, 6 months ago

I solved it the same way, but with so many options, none of these seemed the wrong choice. So, I clicked view solution.

A Former Brilliant Member - 5 years, 6 months ago

Never give up..

Akhil Bansal - 5 years, 6 months ago

The wanted answer is correct. The thing is we usually include a correct figure to be chosen for a question like this.

Lu Chee Ket - 5 years, 6 months ago

Arjen Vreugdenhil
Dec 14, 2015

The decrease of voltage in an RC-circuit is a process of exponential decay. The characteristic time (or relaxation time ), $\tau$ , of an RC-circuit is the time in which the voltage decreases by a factor $e$ . This time is equal is $\tau = RC$ .

In this case the voltage is divided by two. This corresponds to $\ln 2 = 0.693$ characteristic times. Thus the characteristic time for this circuit is $\tau = \frac{5}{0.693} = 7.21\ \text{s}.$ Therefore $\tau = RC \ \ \ \therefore\ \ \ \ C = \frac{\tau}R = \frac{7.21\ \text{s}}{1\times10^6\ \Omega} = 7.21\times10^{-6}\ \text{F}.$ This is not one of the listed solutions, so $\boxed{\text{none of the above}}.$

Ashish Menon
Dec 14, 2015

I did not know it. I just chose none of these. Haha!!

Can you handle this

3 solutions

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