CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 13 of 14

Marks: +1, -0

SECTION - C

An alternating current

I=14 sin \;(100 π t)

A passes through a series combination of a resistor of 30

Ω

and an inductor of

(\;{2}/{5 π})

H. Taking

{√{2}}=1.4

, calculate the
(i) rms value of the voltage drops across the resistor and the inductor, and
(ii) power factor of the circuit.

Solution: 👈: Video Solution

(i)

V_{\;\text "rms "\;}

drop across resistor,

R

\;=I_{ {\rms}} × R

\;=\;{14}/{{√{2}}} × 30

\;=294 {\V}

{\V}_{ {\rms}} \;\text " drop across inductor, "\; {\L}

=I_{\rms} × ω L

=(14 \/ {√{ }} 2) ×(100 π) ×(2 \/ 5 π)

=392 {\V}

(ii) Power factor,

R \/ Z

\;\;{R}/{Z}=\;{R}/{{√{R^2+X_L^2}}}

\;\;{R}/{Z}=\;{30}/{√{30^2+ (100 π × \;{2}/{5 π}) ^2}}

\;=\;{30}/{{√{30^2+40^2}}}

\;=\;{30}/{50}

\;=0.6

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