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Physics
For the circuit shown in figure, the ratio of the voltage across the resistor to across the inductor when the current in the circuit is 2 A, is
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Q. For the circuit shown in figure, the ratio of the voltage across the resistor to across the inductor when the current in the circuit is $2\, A$, is
JIPMER
JIPMER 2013
Alternating Current
A
0.4
16%
B
1.6
33%
C
0.8
42%
D
1.8
9%
Solution:
When current through circuit is 2 A, the voltage drop across the resistor is,
$V_R = IR = 2 \times 8 = 16\, V$
Applying $KVL$, we get $V_L = 36 - 16 = 20\, V$
So, the required ratio is $ \frac{V_R}{V_L} = \frac{16 \, V}{20 \, V} = 0.8$