An inductor coil and a capacitor and a resistance of 5 Ω are connected in series to an a.c. source of rms voltage 30 V . When the frequency of the source is varied, a maximum r.m.s. current of 5 A is observed. If this inductor is connected in parallel with a resistance 5 Ω to a battery of emf 25 V and internal resistance 2.0 Ω , the current drawn from the battery is

Question

An inductor coil and a capacitor and a resistance of 5 Ω are connected in series to an a.c. source of rms voltage 30 V . When the frequency of the source is varied, a maximum r.m.s. current of 5 A is observed. If this inductor is connected in parallel with a resistance 5 Ω to a battery of emf 25 V and internal resistance 2.0 Ω , the current drawn from the battery is (A) (1 5 0/1 7)A (B) 2.5A (C) (1 2 5/3 3)A (D) (1 2 5/9 9)A

Tardigrade Admin · Accepted Answer

Let resistance of inductor be μ textI textmax=(3 0/5 + μ )=5⇒ μ =1Ω Solution

Now, textI = (2 5/(5)6 + 2) Solution

=(1 5 0/1 7) textA

$\frac{150}{17} A$

$2.5 A$

$\frac{125}{33} A$

$\frac{125}{99} A$

Let resistance of inductor be $μ$
$I_{max} = \frac{30}{5 + μ} = 5 \Rightarrow μ = 1Ω$

Now, $I = \frac{25}{\frac{5}{6} + 2}$

$= \frac{150}{17} A$

17150​A

2.5A

33125​A

99125​A

Let resistance of inductor be μ Imax​=5+μ30​=5⇒μ=1Ω Now, I=65​+225​ =17150​A

$\frac{150}{17} A$

$2.5 A$

$\frac{125}{33} A$

$\frac{125}{99} A$

Let resistance of inductor be $μ$
$I_{max} = \frac{30}{5 + μ} = 5 \Rightarrow μ = 1Ω$

Now, $I = \frac{25}{\frac{5}{6} + 2}$

$= \frac{150}{17} A$