A constant voltage is applied to a series R₁ 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
The Correct Answer and Explanation is :
In an RL circuit, the time constant (τ) is defined as the ratio of inductance (L) to resistance (R):
[ \tau = \frac{L}{R} ]
This parameter characterizes how quickly the circuit responds to changes in voltage.
Calculating the Time Constant (τ):
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 τ.
The formula for τ using the two-point method is:
[ \tau = \frac{(t_2 – t_1)}{\ln\left(\frac{V_1}{V_2}\right)} ]
Where:
- ( t_1 = 3.46 \, \text{ms} )
- ( t_2 = 25 \, \text{ms} )
- ( V_1 = 20 \, \text{V} )
- ( V_2 = 5 \, \text{V} )
Substituting the values:
[ \tau = \frac{(25 \, \text{ms} – 3.46 \, \text{ms})}{\ln\left(\frac{20}{5}\right)} ]
[ \tau = \frac{21.54 \, \text{ms}}{\ln(4)} ]
[ \tau \approx \frac{21.54 \, \text{ms}}{1.386} ]
[ \tau \approx 15.54 \, \text{ms} ]
Determining the Resistance (R):
With τ and L known, we can calculate R:
[ R = \frac{L}{\tau} ]
Given that L = 2 H:
[ R = \frac{2 \, \text{H}}{15.54 \times 10^{-3} \, \text{s}} ]
[ R \approx 128.7 \, \Omega ]
Calculating the Current (i₂) at t = 34.7 ms:
After the switch moves to position 2 at t = 0, the circuit’s time constant becomes τ = 25 ms.
The current in an RL circuit after a switch is closed is given by:
[ i(t) = I_0 \left(1 – e^{-\frac{t}{\tau}}\right) ]
Where:
- ( I_0 ) is the final steady-state current
- ( t ) is the time
- ( \tau ) is the time constant
The final current ( I_0 ) is determined by the applied voltage and the total resistance in the circuit.
Assuming the applied voltage is 6 V and the total resistance is 15 Ω:
[ I_0 = \frac{6 \, \text{V}}{15 \, \Omega} = 0.4 \, \text{A} ]
Now, calculate the current at t = 34.7 ms:
[ i(34.7 \, \text{ms}) = 0.4 \, \text{A} \left(1 – e^{-\frac{34.7 \times 10^{-3} \, \text{s}}{25 \times 10^{-3} \, \text{s}}}\right) ]
[ i(34.7 \, \text{ms}) = 0.4 \, \text{A} \left(1 – e^{-1.388}\right) ]
[ i(34.7 \, \text{ms}) \approx 0.4 \, \text{A} \times (1 – 0.249) ]
[ i(34.7 \, \text{ms}) \approx 0.4 \, \text{A} \times 0.751 ]
[ i(34.7 \, \text{ms}) \approx 0.3 \, \text{A} ]
Therefore, the current at t = 34.7 ms is approximately 0.3 A.
For a visual explanation of RL circuits and their time constants, you might find the following video helpful:
video20.4 RL Circuits | General Physicsturn0search2