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

Question

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 (A) 0.4 (B) 1.6 (C) 0.8 (D) 1.8

Tardigrade Admin · Accepted Answer

When current through circuit is 2 A, the voltage drop across the resistor is, VR = IR = 2 × 8 = 16 V Applying KVL, we get VL = 36 - 16 = 20 V So, the required ratio is (VR/VL) = (16 V/20 V) = 0.8

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

0.4

1.6

0.8

1.8

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

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 2A, is

0.4

1.6

0.8

1.8

When current through circuit is 2 A, the voltage drop across the resistor is, VR​=IR=2×8=16V Applying KVL, we get VL​=36−16=20V So, the required ratio is VL​VR​​=20V16V​=0.8

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

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