(D) The phase angle $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.Given $\phi = 45^{\circ}$,$R = 80 \O

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(D) The phase angle $\phi$ between voltage and current in an $RL$ circuit is given by $\tan \phi = \frac{X_L}{R}$.Given $\phi = 45^{\circ}$,$R = 80 \Omega$,and $f = 480 \text{ Hz}$.Since $\tan 45^{\circ} = 1$,we have $1 = \frac{X_L}{R}$,which implies $X_L = R$.Substituting $X_L = 2 \pi f L$,we get $2 \pi f L = R$.Solving for $L$: $L = \frac{R}{2 \pi f}$.Substituting the values: $L = \frac{80}{2 \pi \times 480} = \frac{80}{960 \pi} = \frac{1}{12 \pi} \text{ H}$.

Class 12 Physics Alternating Current RL, RC and LC AC Circuits

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An inductance coil has a resistance of $80 \Omega$. When an $AC$ signal of frequency $480 \text{ Hz}$ is applied to the coil,the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]$

MHT CET 2024 All MHT CET

A
$\frac{1}{24 \pi}$
B
$\frac{\pi}{20}$
C
$\frac{\pi}{40}$
D
$\frac{1}{12 \pi}$

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An inductance coil has a resistance of $80 \Omega$. When an $AC$ signal of frequency $480 \text{ Hz}$ is applied to the coil,the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=1 / \sqrt{2}\right]$

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In an $ac$ circuit,the potential difference across an inductance and resistance joined in series are respectively $16\, V$ and $20\, V$. The total potential difference across the circuit is.......$V$

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An $AC$ voltage $V = 5 \cos(1000t) \text{ V}$ is applied to an $L-R$ circuit of inductance $3 \text{ mH}$ and resistance $4 \text{ } \Omega$. The value of maximum current in the circuit is . . . . . . $A$.

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