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Electromagnetic Waves question

2022 · 25 Jul · Shift 1 · Q54

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Electromagnetic Waves
2022 · 25 Jul · Shift 1 · Q54

Electromagnetic Waves question

2022 · 25 Jul · Shift 1 · Q54

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1

The rms value of conduction current in a parallel plate capacitor is

6.9 \,\mu \mathrm{A}

. The capacity of this capacitor, if it is connected to

230 \mathrm{~V}

ac supply with an angular frequency of

600 \,\mathrm{rad} / \mathrm{s}

, will be :

A
5 pF
B
50 pF
C
100 pF
D
200 pF

View written solutionFree

Correct answer: B

Use the rms current relation for a capacitor

For a capacitor in AC circuit,

m rms} = V_{ m rms}\,\omega C$$ Given: - $I_{ m rms} = 6.9\,\mu A = 6.9\times 10^{-6}\,A$ - $V_{ m rms} = 230\,V$ - $\omega = 600\,\text{rad/s}$ We need to find $C$. 2. **Rearrange the formula** $$C = \frac{I_{ m rms}}{\omega V_{ m rms}}$$ Substitute values: $$C = \frac{6.9\times 10^{-6}}{600\times 230}$$ 3. **Calculate** First compute the denominator: $$600\times 230 = 138000 = 1.38\times 10^5$$ So, $$C = \frac{6.9\times 10^{-6}}{1.38\times 10^5}$$ $$C = 5\times 10^{-11}\,F$$ 4. **Convert into pF** Since $$1\,pF = 10^{-12}\,F$$ Therefore, $$C = 5\times 10^{-11}\,F = 50\times 10^{-12}\,F = 50\,pF$$ 5. **Check options** - A: $5\,pF$ - B: $50\,pF$ ✅ - C: $100\,pF$ - D: $200\,pF$ Hence the correct option is **B**.

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