Electromagnetic Induction and Alternating Current 6 Question 2

2. An alternating voltage $V(t)=220 \sin 100 \pi t$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is

(a) $5 ms$

(b) $2.2 ms$

(c) $7.2 ms$

(d) $3.3 ms$

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Answer:

Correct Answer: 2. (d)

Solution:

  1. In an $AC$ resistive circuit, current and voltage are in phase.

$$ \begin{array}{ll} \text { So, } & I=\frac{V}{R} \\ \Rightarrow & I=\frac{220}{50} \sin (100 \pi t) \end{array} $$

$\therefore$ Time period of one complete cycle of current is

$$ T=\frac{2 \pi}{\omega}=\frac{2 \pi}{100 \pi}=\frac{1}{50} s $$

So, current reaches its maximum value at

$$ t _1=\frac{T}{4}=\frac{1}{200} s $$

When current is half of its maximum value, then from Eq. (i), we have

$$ I=\frac{I _{\max }}{2}=I _{\max } \sin \left(100 \pi t _2\right) $$

$\Rightarrow \sin \left(100 \pi t _2\right)=\frac{1}{2} \Rightarrow 100 \pi t _2=\frac{5 \pi}{6}$

So, instantaneous time at which current is half of maximum value is $t _2=\frac{1}{120} s$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$$ \Delta t=t _2-t _1=\frac{1}{120}-\frac{1}{200}=\frac{1}{300} s=3.3 ms $$



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