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7.Alternating Current
hard
One $10\, \,\,V, 60\, W$ bulb is to be connected to $100\, V$ line. The required induction coil has self inductance of value $(f = 50\,Hz)$
A
$0.052 \,H$
B
$2.42\, H$
C
$16.2 \,mH$
D
$1.62\, mH$
Solution
(a)Current through the bulb $i = \frac{P}{V} = \frac{{60}}{{10}} = 6A$
$V = \sqrt {V_R^2 + V_L^2} $
${(100)^2} = {(10)^2} + V_L^2$$ \Rightarrow \,\,{V_L} = 99.5\,\,Volt$
Also ${V_L} = i{X_L} = i \times (2\pi \nu L)$
$ \Rightarrow \,99.5 = 6 \times 2 \times 3.14 \times 50 \times L$$ \Rightarrow \,\,L = 0.052\,\,H$
Standard 12
Physics
