Find the voltage across inductor at $t = \mu \omega $ if an A.C. circuit having supply voltage $E$ consists of a resistor of resistances $3\Omega $ and an inductor of reactance $4\Omega $ as shown in the figure.
A. $2volt$
B. $10volts$
C. Zero
D. $4.8volts$
Answer
Verified
469.5k+ views
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Economics: Engaging Questions & Answers for Success
Trending doubts
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
What are the major means of transport Explain each class 12 social science CBSE
What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?
What is a transformer Explain the principle construction class 12 physics CBSE
Explain sex determination in humans with the help of class 12 biology CBSE