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Standard XII

Physics

Hysteresis

For the circu...

Question

For the circuit shown in figure, the ratio of the voltage across the resistor to the voltage across the inductor, when current in the circuit is 2A is

0.4

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1.6

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0.8

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None of the above

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Solution

The correct option is C 0.8
When current through circuit is 2A, the voltage drop across resistor is $V_{R} = I R = 2 \times 8 = 16 V$
Applying KVL, we get $V_{L} = 36 - 16 = 20 V$
So, required ratio is $\frac{V_{R}}{V_{L}} = \frac{16}{20} = 0.8$

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For the circuit shown in figure, the ratio of the voltage across the resistor to the voltage across the inductor, when current in the circuit is 2A is

The correct option is C 0.8When current through circuit is 2A, the voltage drop across resistor is VR=IR=2×8=16V Applying KVL, we get VL=36−16=20 V So, required ratio is VRVL=1620=0.8

The correct option is C 0.8
When current through circuit is 2A, the voltage drop across resistor is $V_{R} = I R = 2 \times 8 = 16 V$
Applying KVL, we get $V_{L} = 36 - 16 = 20 V$
So, required ratio is $\frac{V_{R}}{V_{L}} = \frac{16}{20} = 0.8$