Explanation Given data: The voltage across capacitor at t = 0.65 s is 5.22 V and The voltage across capacitor at t = 2.21 s is 1 V . Formula used: The expression for the equivalent resistor connected in series is as follows. R e q = R 1 + R 2 + R 3 ..... + R N (1) Here, R e q is the equivalent resistance between two points and R 1 , R 2 , R 3 …, R N are the resistances. The expression for the equivalent resistor connected in parallel is as follows. R e q = 1 R 1 + 1 R 2 + ..... + 1 R N (2) Here, R e q is the equivalent resistance between two points and R 1 , R 2 , R 3 …, R N are the resistances. The expression for the time constant for R L the circuit is as follows. τ = R e q C (3) Here, τ is the time constant, R e q is the equivalent resistance across the capacitor and C is the capacitance of 10 mF capacitor. The expression for the voltage across the capacitor for t > 0 is as follows. v C ( t ) = v C ( 0 ) e − t τ (4) Here, v C ( t ) is the voltage across the capacitor for t > 0 , v C ( 0 ) is the voltage across the capacitor at t = 0 and t is the given time. Calculation: The redrawn circuit diagram is given in Figure 1 for t < 0 . The expression for the voltage across the capacitor for t < 0 is as follows. v C ( 0 − ) = v ( R 1 R 1 + R 2 ) (5) Here, v C ( 0 − ) is the voltage across the capacitor for t < 0 , R 1 and R 2 are the resistances of the R C circuit and v is the voltage of 12.5 V independent source. Substitute 12.5 V for v and 10 Ω for R 2 in equation (5). v C ( 0 − ) = ( 12.5 V ) ( R 1 R 1 + 10 Ω ) The capacitor does not allow sudden change in the voltage. So, the voltage at t = 0 is equal to the voltage at t < 0 . v C ( 0 ) = ( 12.5 V ) ( R 1 R 1 + 10 Ω ) (6) The redrawn circuit diagram is given in Figure 2. 60 Ω and R 1 resistors are connected in series. So, from equation (1), R e q = 60 Ω + R 1 The simplified diagram is shown in Figure 3. Refer to the redrawn Figure 3. Substitute 60 Ω + R 1 for R e q and 10 mF for C in the equation (3). τ = ( 60 Ω + R 1 ) ( 10 mF ) = ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) { ∵ 1 mF = 10 − 3 F } Substitute ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) for τ and ( 12.5 V ) ( R 1 R 1 + 10 Ω ) for v C ( 0 ) , 0 .65 for t and 5.22 V for v C ( t ) in the equation (4). 5.22 V = ( 12.5 V ) ( R 1 R 1 + 10 Ω ) e − 0 .65 ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) Rearrange the above equation. 5.22 V ( 12.5 V ) ( R 1 R 1 + 10 Ω ) = e − 0 .65 ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) ( R 1 + 10 Ω ) 0.4176 V R 1 = e − 0 .65 ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) Solve for R 1 . R 1 = 40 Ω The redrawn circuit diagram is given in Figure 4 for t = 2 . Substitute ( 60 Ω + R 1 ) ( 10 × 10 − 3 F ) for τ and ( 12.5 V ) ( R 1 R 1 + 10 Ω ) for v C ( 0 ) , 2 for t and 40 Ω for R 1 in the equation (4), v C ( 2 ) = ( 12

9th Edition

ISBN: 9780073545516

Author: Hayt, William H. (william Hart), Jr, Kemmerly, Jack E. (jack Ellsworth), Durbin, Steven M.

Publisher: Mcgraw-hill Education,

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Chapter 8, Problem 34E

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Find the value of the resistor R0 and R1 in the circuit.

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Chapter 8, Problem 33E

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