An alternating operatornameemf E=400 sin 100 π t is applied to a circuit containing an inductance of (√2/π) H. If an AC ammeter is connected in the circuit, its reading will be

Question

An alternating operatornameemf E=400 sin 100 π t is applied to a circuit containing an inductance of (√2/π) H. If an AC ammeter is connected in the circuit, its reading will be (A) 4.4 A (B) 1.55 A (C) 2.2 A (D) 3.11 A

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c4907fed57418cbcdcd5776ef910d693-.png" /> Here, L=(√2/π) H ; ω=100 π; XL=ω L=100 π × (√2/π) XL=100 √2 Ω ∵ E0=i0 XL or i0=(440/100 √2) A ; i0=2.2 √2 A Since ammeter is ac ammeter, it will read rms value of current, ∴ ir m s=(2.2 √2/√2)=2.2 A

Q. An alternating $emf E = 400 sin 100 π t$ is applied to a circuit containing an inductance of $\frac{2}{π} H$ . If an AC ammeter is connected in the circuit, its reading will be

4.4 A

1.55 A

2.2 A

3.11 A

Here, $L = \frac{2}{π} H; ω = 100 π$ ;
$X_{L} = ω L = 100 π \times \frac{2}{π}$
$X_{L} = 1002 Ω$
$∵ E_{0} = i_{0} X_{L}$
or $i_{0} = \frac{440}{100 2} A; i_{0} = 2.2 2 A$
Since ammeter is ac ammeter, it will read rms value of current,
$∴ i_{r m s} = \frac{2.2 2}{2} = 2.2 A$

Q. An alternating emfE=400sin100πt is applied to a circuit containing an inductance of π2​​H. If an AC ammeter is connected in the circuit, its reading will be