Question

A 100 volt AC source of frequency 500 Hz is connected to a L–C-R circuit with L = 8.1 mH, C = 12.5 mF and R = 10 W, all connected in series. The potential difference across the resistance is –

• A. 100 V
• B. 200 V
• C. 300 V
• D. 400 V

Answer 1

Answer: (A)

100 V

Answer 2

Z = $\sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}}$

Here X_L = 2pfL = 2 × 3.14 × 500 × (8.1 × 10^–3)

= 25.4 W

and X_C = $\frac{1}{2\pi fC}$ = $\frac{1}{2 \times 3.14 \times 500 \times 12.5 \times 10^{- 6}}$

= 25.4 W

\ Z = $\sqrt{(10)^{2} + (25.4 - 25.4)^{2}}$ = 10 W

Now i_rms = $\frac{E_{rms}}{Z}$ = $\frac{100}{10}$ = 10 A

\ V_R = i_rms × R

= 10 × 10

= 100 V