
The current in a circuit varies with time as $I = 2\sqrt t $. The RMS value of the current for the interval $t = 2s$ to $t = 4s$ is
A) $\sqrt 3 \,{\text{A}}$
B) $2\sqrt 3 \,{\text{A}}$
C) $\dfrac{{\sqrt 3 }}{2}\,{\text{A}}$
D) $\left( {4 - 2\sqrt 2 } \right){\text{A}}$
Answer
558k+ views
Hint: RMS value stands for the root-mean-square value of the varying current in a given time period. Square the value of current, normalize, and integrate it with respect to time over the specified period and take its square root.
Formula used: ${I_{RMS}} = \sqrt {\dfrac{1}{T}\int {{I^2}dt} } $ where $I$ is the time-varying current in the circuit and $T$ is the interval of time
Complete step by step solution:
We’ve been given that the current in a circuit varies with time as $I = 2\sqrt t $. To find the root-mean-square value of the current in the given interval, we use
$\Rightarrow {I_{RMS}} = \sqrt {\dfrac{1}{T}\int {{I^2}dt} } $
Since the time interval is $t = 2s$ to $t = 4s$, $T = 4 - 2 = 2s$. So,
$\Rightarrow {I_{RMS}} = \sqrt {\dfrac{1}{2}\int\limits_{t = 2}^4 {{{\left( {2\sqrt t } \right)}^2}dt} } $
$\Rightarrow {I_{RMS}} = \sqrt {2\int\limits_{t = 2}^4 {tdt} } $
On carrying out the integration, we get
$\Rightarrow {I_{RMS}} = \sqrt {2\left. {\dfrac{{{t^2}}}{2}} \right|_{t = 2}^4} $
Placing the values of $t = 2s$ and $t = 4s$ as the upper and lower limits of integration, we get
$\Rightarrow {I_{RMS}} = \sqrt {(16 - 4)} = \sqrt {12} $
So the value of${I_{RMS}} = \sqrt {12} = 2\sqrt 3 \,A$ which corresponds to option (B).
Additional Information:
For a sinusoidally current (AC circuit) in the circuit, the RMS value of current is ${I_{RMS}} = \dfrac{{{I_o}}}{{\sqrt 2 }}$ where ${I_o}$ is the maximum current in the circuit. The RMS value for an AC circuit holds more information about the circuit than the maximum current in the circuit. If the current in the circuit is constant with time, like in a DC circuit, the RMS value is equal to the constant value.
Note:
The root-mean-square current is a statistical quantity that is often used when dealing with varying currents in the circuit and is used in calculating the average power in the circuit. We can expect the RMS value of the current to lie between the maximum and the minimum value of the current in the circuit i.e. between $2\sqrt 2 $ and $2\sqrt 4 $ which can help us in removing option (A),(C),(D) as the possible answer.
Formula used: ${I_{RMS}} = \sqrt {\dfrac{1}{T}\int {{I^2}dt} } $ where $I$ is the time-varying current in the circuit and $T$ is the interval of time
Complete step by step solution:
We’ve been given that the current in a circuit varies with time as $I = 2\sqrt t $. To find the root-mean-square value of the current in the given interval, we use
$\Rightarrow {I_{RMS}} = \sqrt {\dfrac{1}{T}\int {{I^2}dt} } $
Since the time interval is $t = 2s$ to $t = 4s$, $T = 4 - 2 = 2s$. So,
$\Rightarrow {I_{RMS}} = \sqrt {\dfrac{1}{2}\int\limits_{t = 2}^4 {{{\left( {2\sqrt t } \right)}^2}dt} } $
$\Rightarrow {I_{RMS}} = \sqrt {2\int\limits_{t = 2}^4 {tdt} } $
On carrying out the integration, we get
$\Rightarrow {I_{RMS}} = \sqrt {2\left. {\dfrac{{{t^2}}}{2}} \right|_{t = 2}^4} $
Placing the values of $t = 2s$ and $t = 4s$ as the upper and lower limits of integration, we get
$\Rightarrow {I_{RMS}} = \sqrt {(16 - 4)} = \sqrt {12} $
So the value of${I_{RMS}} = \sqrt {12} = 2\sqrt 3 \,A$ which corresponds to option (B).
Additional Information:
For a sinusoidally current (AC circuit) in the circuit, the RMS value of current is ${I_{RMS}} = \dfrac{{{I_o}}}{{\sqrt 2 }}$ where ${I_o}$ is the maximum current in the circuit. The RMS value for an AC circuit holds more information about the circuit than the maximum current in the circuit. If the current in the circuit is constant with time, like in a DC circuit, the RMS value is equal to the constant value.
Note:
The root-mean-square current is a statistical quantity that is often used when dealing with varying currents in the circuit and is used in calculating the average power in the circuit. We can expect the RMS value of the current to lie between the maximum and the minimum value of the current in the circuit i.e. between $2\sqrt 2 $ and $2\sqrt 4 $ which can help us in removing option (A),(C),(D) as the possible answer.
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