A 300 Ω resistor is connected in series with a parallel-plate capacitor across the terminals of a 50.0 Hz ac generator. When the gap between the plates is empty, its capacitance is 70/22 μ F. The ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica of dielectric constant k = 5.0 is inserted between the plates, is equal to

Question

A 300 Ω resistor is connected in series with a parallel-plate capacitor across the terminals of a 50.0 Hz ac generator. When the gap between the plates is empty, its capacitance is 70/22 μ F. The ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica of dielectric constant k = 5.0 is inserted between the plates, is equal to (A) 0.1 (B) 0.3 (C) 0.6 (D) 2.9

Tardigrade Admin · Accepted Answer

irms = (E0/√2Z) ⇒ (irms,1/irms, 2) = (Z2/Z1) = (√R2 + ((1/ω kC))2/√R2 + ((1)ω C)2) Solving: ⇒ (irms, 1/irms, 2) = 0.3

$0.1$

$0.3$

$0.6$

$2.9$

$i_{r m s} = \frac{E _{0}}{2 Z}$
$\Rightarrow \frac{i _{r m s, 1}}{i _{r m s, 2}} = \frac{Z _{2}}{Z _{1}}$
$= \frac{R ^{2} + ( \frac{1}{ωk C} ) ^{2}}{R ^{2} + ( \frac{1}{ω C} ) ^{2}}$
Solving : $\Rightarrow \frac{i _{r m s, 1}}{i _{r m s, 2}} = 0.3$

0.1

0.3

0.6

2.9

irms​=2​ZE0​​ ⇒irms,2​irms,1​​=Z1​Z2​​ =R2+(ωC1​)2​R2+(ωkC1​)2​​ Solving : ⇒irms,2​irms,1​​=0.3

$0.1$

$0.3$

$0.6$

$2.9$

$i_{r m s} = \frac{E _{0}}{2 Z}$
$\Rightarrow \frac{i _{r m s, 1}}{i _{r m s, 2}} = \frac{Z _{2}}{Z _{1}}$
$= \frac{R ^{2} + ( \frac{1}{ωk C} ) ^{2}}{R ^{2} + ( \frac{1}{ω C} ) ^{2}}$
Solving : $\Rightarrow \frac{i _{r m s, 1}}{i _{r m s, 2}} = 0.3$