A condenser of capacity 1$\mu F$ and resistance 0.5 megohm are connected in series with a D.C. supply of 2V. The time constant of the circuit is?
A. 0.25 sec
B. 0.5 sec
C. 1 sec
D. 2 sec
Answer
642.9k+ views
Hint: Time constant denotes the time at which the capacitor is charged to the value of 0.63 times the maximum charge that can be stored on the capacitor from zero. The value of time constant ($\tau $) is equal to the product of the capacitance (C) of the capacitor and the resistance (R) of the circuit i.e. $\tau $ = RC.
Formula used:
$q={{q}_{o}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$
$\tau $ = RC
Complete step by step answer:
When a capacitor and resistance are connected in series with a D.C supply, the charge on the capacitor increases exponentially with time. This means that when we plot the graph of charge on the capacitor v/s time, we will get an exponential curve.
Charge on the capacitor as a function of time in this circuit is given as $q={{q}_{o}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$ ……(i).
q is the charge on the capacitor at a given time t, ${{q}_{o}}$ is the maximum amount of charge stored on the capacitor and $\tau $ is the time constant of the circuit.
The value of time constant ($\tau $) is equal to the product of the capacitance (C) of the capacitor and the resistance (R) of the circuit i.e. $\tau $ = RC.
If we consider the time at which the charge on the capacitor is equal to zero, then time constant is equal to the time at which the charge stored on the capacitor is equal to 63% of the maximum charge (${{q}_{o}}$) that can be stored on the capacitor, i.e. q = 0.63${{q}_{o}}$.
The SI unit of time constant is seconds.
Let us find the value of the time constant for the given circuit.
It is given that the capacitance of the capacitor is equal to 1$\mu F$ and the resistance of the circuit is equal to 0.5 megohm. And we know that the value of time constant is equal to $\tau $ = RC.
Therefore, $\tau =\left( 0.5\times {{10}^{6}} \right)\left( 1\times {{10}^{-6}} \right)=0.5\sec $
Hence, the correct answer is B.
Note: The unit of time constant can be found by calculating the dimensional formula of RC. However, there is another way to calculate the unit of time constant.
If we look at equation (i), e is raised to a power of $\dfrac{-t}{\tau }$.
However, powers are just real numbers and do not have any dimension and units. In other words, the magnitude of physical quantity can be raised to a number but a number cannot be raised to the magnitude of a physical quantity. It just does not make any sense.
Therefore, the power $\dfrac{-t}{\tau }$ must be dimensionless. For this, the time constant must have the dimension of time.
Formula used:
$q={{q}_{o}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$
$\tau $ = RC
Complete step by step answer:
When a capacitor and resistance are connected in series with a D.C supply, the charge on the capacitor increases exponentially with time. This means that when we plot the graph of charge on the capacitor v/s time, we will get an exponential curve.
Charge on the capacitor as a function of time in this circuit is given as $q={{q}_{o}}\left( 1-{{e}^{\dfrac{-t}{\tau }}} \right)$ ……(i).
q is the charge on the capacitor at a given time t, ${{q}_{o}}$ is the maximum amount of charge stored on the capacitor and $\tau $ is the time constant of the circuit.
The value of time constant ($\tau $) is equal to the product of the capacitance (C) of the capacitor and the resistance (R) of the circuit i.e. $\tau $ = RC.
If we consider the time at which the charge on the capacitor is equal to zero, then time constant is equal to the time at which the charge stored on the capacitor is equal to 63% of the maximum charge (${{q}_{o}}$) that can be stored on the capacitor, i.e. q = 0.63${{q}_{o}}$.
The SI unit of time constant is seconds.
Let us find the value of the time constant for the given circuit.
It is given that the capacitance of the capacitor is equal to 1$\mu F$ and the resistance of the circuit is equal to 0.5 megohm. And we know that the value of time constant is equal to $\tau $ = RC.
Therefore, $\tau =\left( 0.5\times {{10}^{6}} \right)\left( 1\times {{10}^{-6}} \right)=0.5\sec $
Hence, the correct answer is B.
Note: The unit of time constant can be found by calculating the dimensional formula of RC. However, there is another way to calculate the unit of time constant.
If we look at equation (i), e is raised to a power of $\dfrac{-t}{\tau }$.
However, powers are just real numbers and do not have any dimension and units. In other words, the magnitude of physical quantity can be raised to a number but a number cannot be raised to the magnitude of a physical quantity. It just does not make any sense.
Therefore, the power $\dfrac{-t}{\tau }$ must be dimensionless. For this, the time constant must have the dimension of time.
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