Question

A capacitor is connected to a 12V battery through a resistance of 10Ohm. It is found that the potential difference across the capacitor rises to 4V in 1 micro seconds . Find the capacitance of the capacitor.

Answer 1

Answer: $C = 0.025\mu F$

Explanation:

Given that,

Potential difference $V_{0} = 12 V$

The potential difference across the capacitor rise to 4 V.

Time  $t = 1\mu sec$

The equation of the charge is

$Q = Q_{0}(1-e^{\dfrac{-t}{RC}})$....(I)

We know,

$Q = VC$

Now, put the value of Q in equation (I)

$CV= CV_{0}(1-e^{\dfrac{-t}{RC}})$

After t = 1 micro sec

$4C=12C(1-e^{\dfrac{-1\times10^{-6}sec}{100\Omega\times C}})$

$1-e^{(\dfrac{-1\times10^{-6}sec}{100\times C})} = \dfrac{1}{3}$

$e^{(\dfrac{-1\times10^{-6}sec}{100\times C})} = \dfrac{2}{3}$

$\dfrac{-1\times10^{-6}sec}{100\times C} = ln\dfrac{2}{3}$

$\dfrac{-1\times10^{-6}sec}{100\times C} = -0.4055$

$C = 0.025\mu F$

Hence, the capacitance of the capacitor is $C = 0.025\mu F$.

Answer 2

According to the given data, charge on a capacitor is calculated by

$Q=Q_{ 0 }(1-{ e }^{ \frac { -t }{ RC } })$

Given data states that the potential difference is 4 V in 1 microseconds therefore the maximum potential difference be 12 V.

Thereby,

$Q=Q_{ 0 }(1-e^{ \left( \frac { -t }{ RC } \right) })$

$\Rightarrow CV=CV_{ 0 }(1-e^{ \left( \frac { -t }{ RC } \right) })$

$\Rightarrow After\quad t=10^{ -6 }seconds$

$\Rightarrow 4C=12C(1-e^{ \left( \frac { -{ 10 }^{ -6 } }{ RC } \right) })$

$\Rightarrow\frac { 1 }{ 3 } =(1-e^{ \left( \frac { -{ 10 }^{ -6 } }{ RC } \right) })$

$\Rightarrow -e^{ \left( \frac { -{ 10 }^{ -6 } }{ RC } \right)} = \frac { 2 }{ 3 }$

$\Rightarrow -\frac { -{ 10 }^{ -6 } }{ RC } = ln(\frac { 2 }{ 3 } )$

$\Rightarrow \frac { -{ 10 }^{ -6 } }{ RC } = -ln(\frac { 2 }{ 3 } )$

$\Rightarrow\frac { -{ 10 }^{ -6 } }{ RC } =ln(\frac { 3 }{ 2 } )$

$\Rightarrow \frac { -{ 10 }^{ -6 } }{ RC } =0.4054$

$R= 10ohm$

$\Rightarrow RC=-2.47 \times 10^{ -6 }$

$\Rightarrow C= -2.47 \times 10^{ \left( -5 \right) }$

Neglecting\quad '-'\quad sign.

$\Rightarrow C=2.47\times 10^{ -5 }F$