{"id":189433,"date":"2025-02-09T07:37:06","date_gmt":"2025-02-09T07:37:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189433"},"modified":"2025-02-09T07:37:09","modified_gmt":"2025-02-09T07:37:09","slug":"a-constant-voltage-is-applied-to-a-series-r%e2%82%81-circuit-at-t-0-the-voltage-across-the-inductance-is-20-v-at-3-46-ms-and-5-v-at-25-ms","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/09\/a-constant-voltage-is-applied-to-a-series-r%e2%82%81-circuit-at-t-0-the-voltage-across-the-inductance-is-20-v-at-3-46-ms-and-5-v-at-25-ms\/","title":{"rendered":"A constant voltage is applied to a series R\u2081 circuit at t = 0 The voltage across the inductance is 20 V at 3.46 ms and 5 V at 25 ms"},"content":{"rendered":"\n

A constant voltage is applied to a series R\u2081 circuit at t = 0 The voltage across the inductance is 20 V at 3.46 ms and 5 V at 25 ms. Obtain R if L = 2 Using the two-point method of Section 7-6. tau= t 2 -t 1 ln nu 1 -ln nu 2 = (25 – 3.46)\/(ln(20) – ln(5)) = 15.54ms R = L\/tau = 2\/(15.54 * 10 ^ – 1) = 128.7Omega and so
In the circuit shown in Fig. 7-30, the switch is moved to position 2 at t = 0 Obtain the current i 2 at t = 34.7 ms. After the switching, the three inductances have the equivalent value of L eq = 10 6 + 5(10) 15 =5 H
Then tau = 5\/200 = 25 , and so, with t in ms, i = 6e ^ ((- t) \/ 25) * (A) i_{2}(34.7) = 2e ^ (- 34.7\/29) i_{2} = (5\/15) * i = 2e ^ ((- i) \/ 25) * (A) A = 0.5A<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

In an RL circuit, the time constant (\u03c4) is defined as the ratio of inductance (L) to resistance (R):<\/p>\n\n\n\n

[ \\tau = \\frac{L}{R} ]<\/p>\n\n\n\n

This parameter characterizes how quickly the circuit responds to changes in voltage.<\/p>\n\n\n\n

Calculating the Time Constant (\u03c4):<\/strong><\/p>\n\n\n\n

Given that the voltage across the inductor decreases from 20 V at 3.46 ms to 5 V at 25 ms, we can use the two-point method to determine \u03c4.<\/p>\n\n\n\n

The formula for \u03c4 using the two-point method is:<\/p>\n\n\n\n

[ \\tau = \\frac{(t_2 – t_1)}{\\ln\\left(\\frac{V_1}{V_2}\\right)} ]<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n