Two inductors L1 inductance 1 mH- internal resistance 3- and L2 inductance 2 mH- internal resistance 4- and a resistor R resistance 12- are all connected in parallel across a 5V battery- The circuit its switched on at time t-0- The ratio of the maximum to the minimum current- l max - l min drawn from the battery is

I= V R = 5 12 A I= V R _ Net = 5 3 × 2= 10 3 A I _ max I _ min = 10× 12 3× 5 =8 ...

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Byju's Answer

Standard VIII

Physics

Capacitors

Two inductors...

Question

Two inductors $L_{1}$ (inductance $1 m H$ , internal resistance $3 Ω$ ) and $L_{2}$ (inductance $2 m H$ , internal resistance $4 Ω$ ) and a resistor $R$ (resistance $12 Ω$ ) are all connected in parallel across a $5 V$ battery. The circuit its switched on at time $t = 0$ . The ratio of the maximum to the minimum current. $(l_{m a x} / l_{m i n})$ drawn from the battery is

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Solution

$I = \frac{V}{R} = \frac{5}{12} A I = \frac{V}{R_{N e t}} = \frac{5}{3} \times 2 = \frac{10}{3} A \frac{I_{m a x}}{I_{m i n}} = \frac{10 \times 12}{3 \times 5} = 8$

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I=VR=512AI=VRNet=53×2=103AImaxImin=10×123×5=8

$I = \frac{V}{R} = \frac{5}{12} A I = \frac{V}{R_{N e t}} = \frac{5}{3} \times 2 = \frac{10}{3} A \frac{I_{m a x}}{I_{m i n}} = \frac{10 \times 12}{3 \times 5} = 8$