Ncert Exercises & Solutions

7.15 A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

Given,

Capacitance, 𝐶 = 100 μF

Resistance, 𝑅 = 40 Ω,

rms voltage, 𝑉rms = 110 V

frequency, 𝑓 = 60 Hz

(a) For a capacitive circuit, the maximum current-

$\begin{array}{l} I_{m} = \frac{\sqrt{2} \times V_{r m s}}{\sqrt{X_{C}^{2} + R^{2}}} \\ I_{m} = \frac{\sqrt{2} \times 110 V}{\sqrt{(\frac{1}{2 π \times 60 H z \times 100 μ F} Ω)^{2} + (40 Ω)^{2}}} = 3.23 A \end{array}$

(b) In 𝑅C circuit,
If, 𝐼 = 𝐼𝑚 cos 𝜔t then,

$V = V_{m} \cos (ω t - ϕ)$

current is maximum at t = 0 whereas, voltage is maximum at 𝑡 = 𝜙/𝜔,

Therefore, the time lag between the maximum voltage and maximum
current = 𝜙/𝜔,

$\begin{array}{l} Also, \tan ϕ = \frac{1}{ω C R} = \frac{1}{2 π \times 60} \frac{1}{12 \times 100 μ F \times 100 Ω} = 0.66 \\ ϕ = \tan^{- 1} 0.66 = {33.56}^{\circ} = \frac{33.56 π}{180} \\ time lag = \frac{ϕ}{ω} = \frac{33.56 π}{180 \times 2 π \times 60} = 1.55 m s \end{array}$

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