In the circuit of the figure, the source frequency is ω=2000 rad s -1. The current in the circuit will be

Question

(A) 2 A (B) 3.3 A (C) 2 / √5 A (D) √5 A

Tardigrade Admin · Accepted Answer

XL=L ω=5 × 10-3 × 2000=10 Ω XC=(1/C ω)=(1/50 × 10-6 × 2000) =(100/10) Ω=10 Ω Since X L =X C , therefore Z =R=5.9+0.10+4=10 Ω I v =(E v /Z)=(20/10) A =2 A

Q. In the circuit of the figure, the source frequency is $ω = 2000 r a d s^{- 1}$ . The current in the circuit will be

$2 A$

$3.3 A$

$2/ 5 A$

$5 A$

$X_{L} = L ω = 5 \times 1 0^{- 3} \times 2000 = 10 Ω$
$X_{C} = \frac{1}{C ω} = \frac{1}{50 \times 1 0 ^{- 6} \times 2000}$
$= \frac{100}{10} Ω = 10 Ω$
Since $X_{L} = X_{C},$ therefore
$Z = R = 5.9 + 0.10 + 4 = 10 Ω$
$I_{v} = \frac{E _{v}}{Z} = \frac{20}{10} A = 2 A$

Q. In the circuit of the figure, the source frequency is ω=2000rads−1. The current in the circuit will be

2A

3.3A

2/5​A

5​A