To light, a 50 ~W, 100 ~V lamp is connected, in series with a capacitor of capacitance \frac50\pi \sqrtx \mu F , with 200 ~V, 50 ~Hz \,AC source. The value of x will be \underline\hspace2cm .

Alternating Current PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q66 with Solution

View written solutionFree

Correct answer: 3

Lamp rating and its resistance

The lamp is rated:

Power $P = 50\,\text{W}$
Voltage $V = 100\,\text{V}$

Assuming the lamp behaves as a resistor at its rated condition,

R = \frac{V^2}{P} = \frac{100^2}{50} = \frac{10000}{50} = 200\,\Omega

So the lamp resistance is

R = 200\,\Omega

Condition for proper lighting

The lamp should glow at its rated voltage, so the voltage across the lamp must be

V_R = 100\,\text{V}

Since the lamp is resistive,

I = \frac{V_R}{R} = \frac{100}{200} = 0.5\,\text{A}

Thus the circuit current must be

I = 0.5\,\text{A}

Total impedance of series circuit

The AC source is

V = 200\,\text{V}, \quad f = 50\,\text{Hz}

So the magnitude of total impedance is

|Z| = \frac{V}{I} = \frac{200}{0.5} = 400\,\Omega

For a series $RC$ circuit,

|Z| = \sqrt{R^2 + X_C^2}

Hence,

400 = \sqrt{200^2 + X_C^2}

Squaring,

160000 = 40000 + X_C^2

X_C^2 = 120000

X_C = 100\sqrt{12} = 200\sqrt{3}\,\Omega

Use capacitive reactance formula

We know

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

Given $f=50\,\text{Hz}$ ,

X_C = \frac{1}{100\pi C}

Also given,

C = \frac{50}{\pi\sqrt{x}}\,\mu\text{F}

Convert to farad:

C = \frac{50}{\pi\sqrt{x}} \times 10^{-6}\,\text{F}

Therefore,

X_C = \frac{1}{100\pi \left(\frac{50}{\pi\sqrt{x}}\times 10^{-6}\right)}

Simplify:

X_C = \frac{1}{\frac{5000}{\sqrt{x}}\times 10^{-6}} = \frac{1}{5\times 10^{-3}/\sqrt{x}} = 200\sqrt{x}\,\Omega

Equate the two values of $X_C$

From impedance calculation,

X_C = 200\sqrt{3}

From capacitance expression,

X_C = 200\sqrt{x}

So,

200\sqrt{x} = 200\sqrt{3}

\sqrt{x} = \sqrt{3}

x = 3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer is also $3$ . So the stored answer matches.

More from Alternating Current

View written solutionFree

Correct answer: 3

Lamp rating and its resistance

The lamp is rated:

Power $P = 50\,\text{W}$
Voltage $V = 100\,\text{V}$

Assuming the lamp behaves as a resistor at its rated condition,

R = \frac{V^2}{P} = \frac{100^2}{50} = \frac{10000}{50} = 200\,\Omega

So the lamp resistance is

R = 200\,\Omega

Condition for proper lighting

The lamp should glow at its rated voltage, so the voltage across the lamp must be

V_R = 100\,\text{V}

Since the lamp is resistive,

I = \frac{V_R}{R} = \frac{100}{200} = 0.5\,\text{A}

Thus the circuit current must be

I = 0.5\,\text{A}

Total impedance of series circuit

The AC source is

V = 200\,\text{V}, \quad f = 50\,\text{Hz}

So the magnitude of total impedance is

|Z| = \frac{V}{I} = \frac{200}{0.5} = 400\,\Omega

For a series $RC$ circuit,

|Z| = \sqrt{R^2 + X_C^2}

Hence,

400 = \sqrt{200^2 + X_C^2}

Squaring,

160000 = 40000 + X_C^2

X_C^2 = 120000

X_C = 100\sqrt{12} = 200\sqrt{3}\,\Omega

Use capacitive reactance formula

We know

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

Given $f=50\,\text{Hz}$ ,

X_C = \frac{1}{100\pi C}

Also given,

C = \frac{50}{\pi\sqrt{x}}\,\mu\text{F}

Convert to farad:

C = \frac{50}{\pi\sqrt{x}} \times 10^{-6}\,\text{F}

Therefore,

X_C = \frac{1}{100\pi \left(\frac{50}{\pi\sqrt{x}}\times 10^{-6}\right)}

Simplify:

X_C = \frac{1}{\frac{5000}{\sqrt{x}}\times 10^{-6}} = \frac{1}{5\times 10^{-3}/\sqrt{x}} = 200\sqrt{x}\,\Omega

Equate the two values of $X_C$

From impedance calculation,

X_C = 200\sqrt{3}

From capacitance expression,

X_C = 200\sqrt{x}

So,

200\sqrt{x} = 200\sqrt{3}

\sqrt{x} = \sqrt{3}

x = 3

Comparison with stored answer

Stored correct answer = $3$

Our derived answer is also $3$ . So the stored answer matches.

Alternating Current question

Alternating Current question

More from Alternating Current

Alternating Current question

Alternating Current question

More from Alternating Current