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7.Alternating Current
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Two resistors are connected in series across a $5\,V\,rms$ source of alternating potential. The potential difference across $6\,\Omega $ resistor is $3\,V$. If $R$ is replaced by a pure inductor $L$ of such magnitude that current remains same, then the potential difference across $L$ is.......$V$
A
$1$
B
$2$
C
$3$
D
$4$
Solution
$\mathrm{V}_{6 \Omega}=3=6 \mathrm{I}_{\mathrm{V}} \therefore \mathrm{I}_{\mathrm{v}}=0.5 \mathrm{\,A}$
$\mathrm{I}_{\mathrm{V}}=\frac{1}{2}=\frac{5}{\sqrt{6^{2}+\mathrm{x}_{\mathrm{L}}^{2}}}, \mathrm{X}_{\mathrm{L}}=8\, \Omega$
Now, $V_{L}=I_{V} \cdot X_{L}=\frac{1}{2} \times 8=4 \,V$
Standard 12
Physics
