A coil of negligible resistance is connected in series with \(90 \, \Omega\) resistor across \(120 \, \text{V}, \, 60 \, \text{Hz}\) supply. A voltmeter reads \(36 \, \text{V}\) across resistance. Inductance of the coil is:
- \(0.76 \, \text{H}\)
- \(2.86 \, \text{H}\)
- \(0.286 \, \text{H}\)
- \(0.91 \, \text{H}\)
The Correct Option is A
Approach Solution - 1
We are given:
\[
36 = I_{\text{rms}} R
\]
Step 1: Substitute the expression for \( I_{\text{rms}} \)
\[
36 = \frac{120}{\sqrt{X_L^2 + R^2}} \times R
\]
Given \( R = 90 \, \Omega \),
\[
36 = \frac{120 \times 90}{\sqrt{X_L^2 + 90^2}}
\]
Step 2: Simplify and solve for \( X_L \)
Rearranging the equation:
\[
\sqrt{X_L^2 + 90^2} = 300
\]
Squaring both sides:
\[
X_L^2 + 90^2 = 300^2
\]
\[
X_L^2 = 90000 - 8100 = 81900
\]
\[
X_L = 286.18 \, \Omega
\]
Step 3: Using \( X_L = \omega L \)
\[
\omega L = 286.18
\]
\[
L = \frac{286.18}{376.8}
\]
\[
L = 0.76 \, \text{H}
\]
Final Answer:
\[
\boxed{L = 0.76 \, \text{H}}
\]
Approach Solution -2
The circuit contains a resistor (\(R\)) and an inductor (\(L\)) in series. The total voltage (\(V\)) is the RMS voltage of the supply. The current in the circuit is:
\[I_{\text{rms}} = \frac{V_R}{R},\]
where \(V_R = 36 \, \text{V}\) and \(R = 90 \, \Omega\). Substituting:
\[I_{\text{rms}} = \frac{36}{90} = 0.4 \, \text{A}.\]
The total impedance of the circuit is:
\[Z = \frac{V}{I_{\text{rms}}} = \frac{120}{0.4} = 300 \, \Omega.\]
The impedance is related to the resistance and inductive reactance by:
\[Z = \sqrt{R^2 + X_L^2}.\]
Substituting \(Z = 300 \, \Omega\) and \(R = 90 \, \Omega\):
\[300 = \sqrt{90^2 + X_L^2}.\]
Squaring both sides:
\[300^2 = 90^2 + X_L^2 \implies X_L^2 = 300^2 - 90^2 = 81900.\]
Thus:
\[X_L = \sqrt{81900} \approx 286.18 \, \Omega.\]
The inductive reactance is given by:
\[X_L = \omega L \quad \text{where} \quad \omega = 2\pi f.\]
For \(f = 60 \, \text{Hz}\):
\[\omega = 2\pi \cdot 60 \approx 376.8 \, \text{rad/s}.\]
Solving for \(L\):
\[L = \frac{X_L}{\omega} = \frac{286.18}{376.8} \approx 0.76 \, \text{H}.\]
Thus, the inductance of the coil is:
\[L = 0.76 \, \text{H}.\]
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Alternating current Questions
- A series \( L, R \) circuit connected with an AC source \( E = (25 \sin 1000 \, t) \, \text{V} \) has a power factor of \( \frac{1}{\sqrt{2}} \). If the source of emf is changed to \( E = (20 \sin 2000 \, t) \, \text{V} \), the new power factor of the circuit will be:
- A series LCR circuit with \( L = \frac{100}{\pi} \) mH, \( C = \frac{10^{-3}}{\pi} \) F and \( R = 10 \, \Omega \) is connected across an AC source of 220 V, 50 Hz supply. The power factor of the circuit would be _____.
- An alternating voltage \( V(t) = 220 \sin 100 \pi t \) volt is applied to a purely resistive load of 50 \( \Omega \). The time taken for the current to rise from half of the peak value to the peak value is:
- A capacitor of reactance \( 4 \sqrt{3} \, \Omega \) and a resistor of resistance \( 4 \, \Omega \) are connected in series with an AC source of peak value \( 8 \sqrt{2} \, \text{V} \). The power dissipation in the circuit is ___________ W.
- In an ac circuit, the instantaneous current is zero,when the instantaneous voltage is maximum. In this case, the source may be connected to :
A. pure inductor.
B. pure capacitor.
C. pure resistor.
D. combination of an inductor and capacitor.
Choose the correct answer from the options given below :
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.