An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 Ω is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

Question

An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 Ω is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be (A) 10 A (B) 8 A (C) 6 √3 A (D) 6 √2 A

Tardigrade Admin · Accepted Answer

Since (100/8) =2π × 50 × L ⇒ L= (1/8π)H Now Xâ²L =2π × 40× (1/8π)=10Ω R = 10Ω Z = √102+102=10√2Ω l = (120/10√2)=6√2A

Q. An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 $Ω$ is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

10 A

8 A

$63$ A

$62$ A

Since $\frac{100}{8} = 2 π \times 50 \times L \Rightarrow L = \frac{1}{8 π} H$
Now $X'_{L} = 2 π \times 40 \times \frac{1}{8 π} = 10Ω$
$R = 10Ω$
$Z = 1 0^{2} + 1 0^{2} = 102 Ω$
$l = \frac{120}{10 2} = 62 A$

Q. An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 Ω is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

10 A

8 A

63​ A

62​ A

Since 8100​=2π×50×L⇒L=8π1​H Now X′L​=2π×40×8π1​=10Ω R=10Ω Z=102+102​=102​Ω l=102​120​=62​A

Q. An ideal choke coil draws a current 8 A when connected to an AC supply of 100 V, 50 Hz. A resistance of 10 $Ω$ is connected in series to choke and then connected to the AC source of 120 V, 40 Hz. The current in the circuit will be

$63$ A

$62$ A

Since $\frac{100}{8} = 2 π \times 50 \times L \Rightarrow L = \frac{1}{8 π} H$
Now $X'_{L} = 2 π \times 40 \times \frac{1}{8 π} = 10Ω$
$R = 10Ω$
$Z = 1 0^{2} + 1 0^{2} = 102 Ω$
$l = \frac{120}{10 2} = 62 A$