A constant voltage is applied to a series R-L circuit by closing the switch. The voltage across the inductor (L=2H) is $20V$ at $t=0$ and drops to $5V$ at $20ms$. The value of R in ohms is:
A. 100ln(2)
B. 100(1-ln2)
C. 100ln4
D. 100(1-ln4)
Answer
642.3k+ views
Hint:The time needed to achieve its full steady state value for the current flowing in the R-L series circuit is equal to about 5 times constants or 5. This time constant is mathematically determined using the formula,$\tau = \dfrac{L}{R}$ where, R is the resistance of the resistor in ohms and L is the value of inductance of the inductor in Henries.
Complete step by step answer:
A R-L series circuit consists essentially of an inductance inductor, L linked in series to a resistance resistor, R. The “R” resistance is the DC resistive value of the wire turns or loops that make up the coil of the inductors.
Given, the value of inductance of the inductor, L=2H
The value remains constant 1/4th in 20ms. Hence, two half lives are equal to 20ms and a half life period is 10ms.
${\tau _{\dfrac{1}{2}}} = 10ms$
The half life period of R-L circuit is given by,
$
{\tau _{\dfrac{1}{2}}} = \ln (2)\dfrac{L}{R} \\
\Rightarrow R = \dfrac{{L \times \ln (2)}}{{{\tau _{\dfrac{1}{2}}}}} \\
\Rightarrow R = \dfrac{{2\ln (2)}}{{10 \times {{10}^{ - 3}}}} \\
\Rightarrow R = 100 \times 2\ln (2) \\
$
Using logarithm identities,
$\log ({a^b}) = b\log (a)$
Using the above identity, we get
$
R = 100 \times 2\ln (2) \\
\Rightarrow R = 100\ln ({2^2}) \\
\therefore R = 100\ln (4)ohms \\
$
Thus, option(C) is correct.
Note: The R-L time constant shows the amount of time it takes to execute 63.2% of the current resulting from a voltage applied across an inductor. For DC power supplies to RF amplifiers, R-L circuits are used, where the inductor is used to transfer DC bias current and block the return of the RF to the power supply.The inductor current in a circuit consisting only of resistors and a single equivalent inductance is represented in the natural response of an R-L circuit. Source-free is the circuit: the response is entirely due to the energy originally contained in the container.
Complete step by step answer:
A R-L series circuit consists essentially of an inductance inductor, L linked in series to a resistance resistor, R. The “R” resistance is the DC resistive value of the wire turns or loops that make up the coil of the inductors.
Given, the value of inductance of the inductor, L=2H
The value remains constant 1/4th in 20ms. Hence, two half lives are equal to 20ms and a half life period is 10ms.
${\tau _{\dfrac{1}{2}}} = 10ms$
The half life period of R-L circuit is given by,
$
{\tau _{\dfrac{1}{2}}} = \ln (2)\dfrac{L}{R} \\
\Rightarrow R = \dfrac{{L \times \ln (2)}}{{{\tau _{\dfrac{1}{2}}}}} \\
\Rightarrow R = \dfrac{{2\ln (2)}}{{10 \times {{10}^{ - 3}}}} \\
\Rightarrow R = 100 \times 2\ln (2) \\
$
Using logarithm identities,
$\log ({a^b}) = b\log (a)$
Using the above identity, we get
$
R = 100 \times 2\ln (2) \\
\Rightarrow R = 100\ln ({2^2}) \\
\therefore R = 100\ln (4)ohms \\
$
Thus, option(C) is correct.
Note: The R-L time constant shows the amount of time it takes to execute 63.2% of the current resulting from a voltage applied across an inductor. For DC power supplies to RF amplifiers, R-L circuits are used, where the inductor is used to transfer DC bias current and block the return of the RF to the power supply.The inductor current in a circuit consisting only of resistors and a single equivalent inductance is represented in the natural response of an R-L circuit. Source-free is the circuit: the response is entirely due to the energy originally contained in the container.
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