Question

In an $RC$ circuit, $R=15k\Omega $, battery $emf=24V$ and time constant $\tau =24\mu s$. Then,
$A)$total capacitance of the circuit is $1.6\times {{10}^{-9}}F$
$B)$time taken by voltage across resistor to reach $16V$ is more than $10\mu s$
$C)$ total capacitance of circuit is $2.8\times {{10}^{-9}}F$
$D)$ time taken by voltage across resistor to reach $16V$ is less than $10\mu s$

Answer 1

Hint: When voltage is supplied to an $RC$ circuit, the capacitor charges up through the resistor. Capacitance of the capacitor used in the $RC$ circuit is determined from the expression for the time constant of the circuit. Expressions for voltage across capacitor as well as voltage across resistor are noted down. These expressions are further used to determine the time taken for the voltage across the resistor to reach a value of $16V$.
Formula used:
$1)\tau =RC$
$2){{V}_{C}}={{V}_{S}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$
$3){{V}_{R}}={{V}_{S}}-{{V}_{C}}$

Complete answer:
We are provided with an $RC$ circuit in which a resistor of resistance $R=15k\Omega $ is used. Voltage supply or battery $emf$ of the circuit is given as $24V$. Also, time constant $\tau $ of the circuit is given as $24\mu s$. With all this information provided, we are required to calculate the capacitance of the capacitor used in the $RC$ as well as the time taken by voltage across the resistor to reach a value of $16V$.

Most of the electric circuits suffer from a time-delay between the input signal and the output signal. This time-delay is generally termed as the time constant of an electric circuit. For an $RC$ circuit, time constant $(\tau )$ is given by
$\tau =RC$
where
$\tau $ is the time constant of an $RC$ circuit
$R$ is the resistance of the resistor used in $RC$ circuit
$C$ is the capacitance of the capacitor used in $RC$ circuit
Let this be equation 1.
Using equation 1, capacitance of the capacitor used in the given $RC$ circuit is given by
$\tau =RC\Rightarrow C=\dfrac{\tau }{R}=\dfrac{24\mu s}{15k\Omega }=\dfrac{24\times {{10}^{-6}}s}{15\times {{10}^{3}}\Omega }=1.6\times {{10}^{-9}}F$
where
$\tau =24\mu s$ is the time constant of the given circuit, as provided in the question
$R=15k\Omega $ is the resistance of the resistor used in the circuit, as provided in the question
Let this be equation 2.
Now, we know that voltage across capacitor in an $RC$ circuit is given by
${{V}_{C}}={{V}_{S}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$
where
${{V}_{C}}$ is the voltage across the capacitor in $RC$ circuit
${{V}_{S}}$ is the supply voltage or the battery $emf$
$\tau $ is the time constant of $RC$ circuit
$t$ is the elapsed time since the application of battery $emf$
Let this be equation 3.
Using equation 3, voltage across capacitor used in the $RC$ given circuit is given by
${{V}_{C}}={{V}_{S}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)=24V\left( 1-{{e}^{\dfrac{-t}{24\mu s}}} \right)=24V-24V\left( {{e}^{\dfrac{-t}{24\mu s}}} \right)$
where
${{V}_{C}}$ is the voltage across the capacitor in the given $RC$ circuit
${{V}_{S}}=24V$ is the battery $emf$
$\tau =24\mu s$ is the time constant
$t$ is the elapsed time since the application of battery $emf$
Let this be equation 4.
Now, we know that voltage across resistor in an $RC$ circuit is given by
${{V}_{R}}={{V}_{S}}-{{V}_{C}}$
where
${{V}_{R}}$ is the voltage across resistor in an $RC$ circuit
${{V}_{S}}$ is the supply voltage or the battery $emf$ of $RC$ circuit
${{V}_{C}}$ is the voltage across the capacitor in the given $RC$ circuit
Let this be equation 5.
From the options provided along with the question, we know that we have to calculate the time required for the voltage of the resistor to reach a value of $16V$. Clearly, we can write the value of ${{V}_{R}}$ in the given circuit as
${{V}_{R}}=16V$
Substituting this value of ${{V}_{R}}$ and equation 4 in equation 5, we have
${{V}_{R}}={{V}_{S}}-{{V}_{C}}\Rightarrow 16V=24V-24V+24V\left( {{e}^{\dfrac{-t}{24\mu s}}} \right)\Rightarrow 16V=24V\left( {{e}^{\dfrac{-t}{24\mu s}}} \right)\Rightarrow {{e}^{\dfrac{t}{24\mu s}}}=1.5$
Taking $\ln $ on both sides, we have
$\ln {{e}^{\dfrac{t}{24\mu s}}}=\ln 1.5\Rightarrow t=24\mu s\times \ln 1.5=24\mu s\times 0.405=9.72\mu s$

Let this be equation 6.
Therefore, from equation 2 and equation 6,
a) total capacitance of the circuit is $1.6\times {{10}^{-9}}F$
b) the elapsed time since the application of battery $emf$ for the voltage across resistor to reach $16V$ is equal to $9.72\mu s$.

So, the correct answer is “Option A and D”.

Note:
Students need to be aware of natural logarithmic function. It should be made clear that
$\ln ({{e}^{x}})={{\log }_{e}}({{e}^{x}})=x$
Thus, derived is the final step of solution given above.
Also, students need not forget to read the options before attempting the solution. Here, the value of voltage across the resistor $({{V}_{R}})$ to be taken, is given in the provided options.