In the circuit shown below, the ac source has voltage V=20 cos (ω t) volts with ω=2000 rad / sec. The amplitude of the current will be nearest to

Question

In the circuit shown below, the ac source has voltage V=20 cos (ω t) volts with ω=2000 rad / sec. The amplitude of the current will be nearest to <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9aeba36ae3b82150f3539e0c2549324d-.png" /> (A) 2 A (B) 3.3 A (C) 2 / √5 A (D) √5 A

Tardigrade Admin · Accepted Answer

R=6+4=10 Ω XL=ω L=2000 × 5 × 10-3=10 Ω XC=(1/ω C)=(1/2000 × 50 × 10-6)=10 Ω ∴ Z =√R2+(XL-XC)2=10 Ω Amplitude of current =i0=(V0/Z)=(20/10)=2 A

Q. In the circuit shown below, the ac source has voltage $V = 20 cos (ω t)$ volts with $ω = 2000 r a d / sec$ . The amplitude of the current will be nearest to

$2 A$

$3.3 A$

$2/ 5 A$

$5 A$

$R = 6 + 4 = 10 Ω$
$X_{L} = ω L = 2000 \times 5 \times 1 0^{- 3} = 10 Ω$
$X_{C} = \frac{1}{ω C} = \frac{1}{2000 \times 50 \times 1 0 ^{- 6}} = 10 Ω$
$∴ Z = R^{2} + (X_{L} - X_{C})^{2} = 10 Ω$
Amplitude of current $= i_{0} = \frac{V _{0}}{Z} = \frac{20}{10} = 2 A$

Q. In the circuit shown below, the ac source has voltage V=20cos(ωt) volts with ω=2000rad/sec. The amplitude of the current will be nearest to

2A

3.3A

2/5​A

5​A