Answer
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Hint Evaluate the total impedance of series inductive – resistor A.C circuit by the expression –
$Z = \sqrt {{R^2} + X_L^2} $
where, $R$ is the resistor of resistance and ${X_L}$ is the inductor of reactance.
Now, put $t = \mu \omega $ in $E = 10\sin \omega t$
Then, $\sin ({\omega ^2} \times {10^{ - 6}})$ will be approximately equal to zero. So, current will be equal to zero.
Complete step-by-step solution:Let $R$ be the resistor of resistance and ${X_L}$ be the inductor of reactance.
So, according to the question, it is given that -
$R = 3\Omega $
${X_L} = 4\Omega $
Now, calculating the impedance for this inductive – resistor circuit so, this calculated by using the formula –
$Z = \sqrt {{R^2} + X_L^2} $
Putting the values of resistance and inductor of reactance in the above equation –
$Z = \sqrt {{3^2} + {4^2}} $
$Z = 5\Omega $
Therefore, the value for impedance of this circuit is $5\Omega $.
Now, it is given that –
$E = 10\sin \omega t$
Also, it is given in equation that, $t = \mu \omega $, therefore, putting this value of $t$ in the equation of $E$, we get
$
E = 10\sin \omega (\mu \omega ) \\
E = 10\sin \mu {\omega ^2} \\
$
We know that, $\mu = {10^{ - 6}}$
$\therefore E = 10\sin ({\omega ^2} \times {10^{ - 6}})$
If we take the value of ${\omega ^2} \times {10^{ - 6}}$ then, we come to know that its value is approximately very low
${\omega ^2} \times {10^{ - 6}} \approx verylow$
Therefore, $\sin ({\omega ^2} \times {10^{ - 6}})$ becomes approximately equal to zero.
$\sin ({\omega ^2} \times {10^{ - 6}}) \approx 0$
Then, current also becomes zero.
So, if we calculate the voltage across inductor, we get that –
${V_L} = i{X_L}$
Because, the current is equal to zero
$\therefore {V_L} = 0$
Therefore, option (C), zero volt is the correct option.
Note:- The R – L circuit can be defined as electrical circuit of elements which includes resistor R and inductance L connected together having source as voltage or as current.
The impedance of series R – L circuit is the combined effect of resistance R and inductive reactance ${X_L}$ of circuit as whole.
$Z = \sqrt {{R^2} + X_L^2} $
where, $R$ is the resistor of resistance and ${X_L}$ is the inductor of reactance.
Now, put $t = \mu \omega $ in $E = 10\sin \omega t$
Then, $\sin ({\omega ^2} \times {10^{ - 6}})$ will be approximately equal to zero. So, current will be equal to zero.
Complete step-by-step solution:Let $R$ be the resistor of resistance and ${X_L}$ be the inductor of reactance.
So, according to the question, it is given that -
$R = 3\Omega $
${X_L} = 4\Omega $
Now, calculating the impedance for this inductive – resistor circuit so, this calculated by using the formula –
$Z = \sqrt {{R^2} + X_L^2} $
Putting the values of resistance and inductor of reactance in the above equation –
$Z = \sqrt {{3^2} + {4^2}} $
$Z = 5\Omega $
Therefore, the value for impedance of this circuit is $5\Omega $.
Now, it is given that –
$E = 10\sin \omega t$
Also, it is given in equation that, $t = \mu \omega $, therefore, putting this value of $t$ in the equation of $E$, we get
$
E = 10\sin \omega (\mu \omega ) \\
E = 10\sin \mu {\omega ^2} \\
$
We know that, $\mu = {10^{ - 6}}$
$\therefore E = 10\sin ({\omega ^2} \times {10^{ - 6}})$
If we take the value of ${\omega ^2} \times {10^{ - 6}}$ then, we come to know that its value is approximately very low
${\omega ^2} \times {10^{ - 6}} \approx verylow$
Therefore, $\sin ({\omega ^2} \times {10^{ - 6}})$ becomes approximately equal to zero.
$\sin ({\omega ^2} \times {10^{ - 6}}) \approx 0$
Then, current also becomes zero.
So, if we calculate the voltage across inductor, we get that –
${V_L} = i{X_L}$
Because, the current is equal to zero
$\therefore {V_L} = 0$
Therefore, option (C), zero volt is the correct option.
Note:- The R – L circuit can be defined as electrical circuit of elements which includes resistor R and inductance L connected together having source as voltage or as current.
The impedance of series R – L circuit is the combined effect of resistance R and inductive reactance ${X_L}$ of circuit as whole.
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