A resistance of 20 ohms is connected to a source of an alternating potential V=220 sin (100 π t). The time taken by the current to change from its peak value to r.m.s. value is

Question

A resistance of 20 ohms is connected to a source of an alternating potential V=220 sin (100 π t). The time taken by the current to change from its peak value to r.m.s. value is (A) 0.2 sec (B) 0.25 sec (C) 25 × 10-3 sec (D) 2.5 × 10-3 sec

Tardigrade Admin · Accepted Answer

Peak value to r.m.s. value means current becomes (1/√2) times. So from i=i0 sin 100 π t ⇒ (1/√2) × i0=i0 sin 100 π t ⇒ sin (π/4)= sin 100 π t ⇒ t=(1/400) sec =2.5 × 10-3 sec

Q. A resistance of $20 o hm s$ is connected to a source of an alternating potential $V = 220 sin (100 π t)$ . The time taken by the current to change from its peak value to r.m.s. value is

$0.2 sec$

$0.25 sec$

$25 \times 1 0^{- 3} sec$

$2.5 \times 1 0^{- 3} sec$

Peak value to r.m.s. value means current becomes $\frac{1}{2}$ times.
So from $i = i_{0} sin 100 π t$
$\Rightarrow \frac{1}{2} \times i_{0} = i_{0} sin 100 π t$
$\Rightarrow sin \frac{π}{4} = sin 100 π t$
$\Rightarrow t = \frac{1}{400} sec = 2.5 \times 1 0^{- 3} sec$

Q. A resistance of 20ohms is connected to a source of an alternating potential V=220sin(100πt). The time taken by the current to change from its peak value to r.m.s. value is

0.2sec

0.25sec

25×10−3sec

2.5×10−3sec