Two inductors (inductance 1 mH, internal resistance ) and (inductance 2 mH, internal resistance ), and a resistor R (resistance ) are all connected in parallel across a 5 V battery. The circuit is switched on at time . The ratio of the maximum to the mi

Question

Two inductors $L_{1}$ (inductance 1 mH, internal resistance $3 Ω$ ) and $L_{2}$ (inductance 2 mH, internal resistance $4 Ω$ ), and a resistor R (resistance $12 Ω$ ) are all connected in parallel across a 5 V battery. The circuit is switched on at time $t = 0$ . The ratio of the maximum to the minimum current $(\frac{I_{\max}}{I_{\min}})$ drawn from the battery is ______ [2016]

Smart Achievers · Accepted Answer

(8)

$At t = 0, I_{\min} = \frac{5}{12}$

At $t = \infty$ ,

$I_{\max} = \frac{5}{R_{eq}} = \frac{5}{3 / 2} = \frac{10}{3}$

$[\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{8}{12}]$

$∴$ $\frac{I_{\max}}{I_{\min}} = \frac{10}{3} \times \frac{12}{5} = 8$

Solution

Ans.
(8)

$At t = 0, I_{\min} = \frac{5}{12}$

At $t = \infty$ ,

$I_{\max} = \frac{5}{R_{eq}} = \frac{5}{3 / 2} = \frac{10}{3}$

$[\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{8}{12}]$

$∴$ $\frac{I_{\max}}{I_{\min}} = \frac{10}{3} \times \frac{12}{5} = 8$

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Solution

Ans. (8) At t=0, Imin=512 At t=∞, Imax=5Req=53/2=103 [1Req=13+14+112=812] ∴ ImaxImin=103×125=8

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Ans.
(8)

$At t = 0, I_{\min} = \frac{5}{12}$

At $t = \infty$ ,

$I_{\max} = \frac{5}{R_{eq}} = \frac{5}{3 / 2} = \frac{10}{3}$

$[\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12} = \frac{8}{12}]$

$∴$ $\frac{I_{\max}}{I_{\min}} = \frac{10}{3} \times \frac{12}{5} = 8$