Solving Simple Circuits Reference: Analog Signal Processing Problems 1,2,3,4,5, by Don Johnson, Rice University,
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1 Problem Set Problem : Solvg Simple Circuits eference: Analog Signal Processg Problems,,,,5, by Don Johnson, ice University, Write the set of equations that govern Circuit A's behavior.. Solve these equations for i : n other words, express this current terms of element and source values by elimatg non-source voltages and currents.. For Circuit B, fd the value for L that results a current of 5 A passg through it. Circuit A Subfigure. Circuit B Subfigure. Figure ) With the current source removed from the circuit (replaced with an open), the current through will be the same as i. V i i 0 ) i i i i i V ( i i 0 ) ) With a current of 5A passg through the load resistor, 0 amps must pass through the 0 ohm resistor. By spection, to halve the amount of current the branch, the resistor value must be doubled. L 0Ω
2 Problem : Equivalent esistance For each of the followg circuits, fd the equivalent resistance usg series and parallel combation rules. circuit a Subfigure. circuit b Subfigure. circuit c Subfigure. circuit d Subfigure. Figure
3 ( ) ( ) 5 5 EQ a) circuit EQ b) circuit ( ) ( ) EQ c) circuit Ω d) circuit EQ Calculate the conductance seen at the termals for circuit (c) terms of each element's conductance. Compare this equivalent conductance formula with the equivalent resistance formula you found for circuit (b). How is the circuit (c) derived from circuit (b)? EQ c) circuit The relationship between circuit b and circuit c pertas to the position and shape of the components. is parallel position circuit b and a series position circuit c. and are parallel positions circuit b and series positions circuit c. is a series position circuit b and a parallel position circuit c.
4 Problem : Superposition Prciple One of the most important consequences of circuit laws is the Superposition Prciple: The current or voltage defed for any element equals the sum of the currents or voltages produced the element by the dependent sources. This Prciple has important consequences simplifyg the calculation of circuit variables multiple source circuits. Figure. For the depicted circuit, fd the dicated current usg any technique you like (you should use the simplest).. You should have found that the current i is a lear combation of the two source values: ic v C i. This result means that we can thk of the current as a superposition of two components, each of which is due to a source. We can fd each component by settg the other sources to zero. Thus, to fd the voltage source component, you can set the current source to zero (an open circuit) and use the usual tricks. To fd the current source component, you would set the voltage source to zero (a short circuit) and fd the resultg current. Calculate the total current i usg the Superposition Prciple. s applyg the Superposition Prciple easier than the technique you used part ()?
5 The current through contributed by the voltage source. tot Ω V Ω V The current through contributed by the current source..5.5 The current through tot.5ω.8ω The total current is: 8.5.5Ω.5.8Ω 8 V
6 Problem : Current and Voltage Divider Use current of voltage divider rules to calculate the dicated circuit variables figure. circuit a Subfigure. circuit c Subfigure. Circuit a) Circuit b) circuit b Subfigure. Figure 5Ω V out 7 s 5t 5s 5t 7Ω (Ω Ω)6Ω ( ) 6 6A Ω Ω Ω 6Ω 6 A
7 Circuit c) Ω tot 0 tot 6Ω 6.956A 0V 7.5 [ 80Ω [ 8Ω ( 5Ω 0Ω Ω) ] 6.956A Ω [ 80Ω [ 8Ω ( Ω 5Ω 0Ω) ] 5Ω 0Ω.978A 0 Ω 5Ω 0Ω 7.5Ω.89A Problem 5: Théven and Mayer-orton Equivalents Fd the Théven and Mayer-orton equivalent circuits for the followg circuits. circuit a Subfigure 5. circuit b Subfigure 5. circuit c Subfigure 5. Figure 5
8 Circuit a) To fd, remove current source from circuit by replacg it with an open. Calculate the total resistance between the load termals. is equal to. (Ω Ω) Ω. Ω To fd, short the load termals and calculate the current that will flow between the load termals. CD can be used to calculate the orton current. Ω Ω A A Ω V.V V. Ω _ V _ 5
9 Circuit b) To fd, remove voltage source from circuit by replacg it with a short. Calculate the total resistance between the load termals. is equal to. (Ω Ω) (Ω Ω) Ω To fd V, remove the load resistance and calculate the voltage that is present at between the load termals. Sce the bridge circuit is balanced, no voltage will be present. V 0.0V V 0A _ V _ 5
10 Circuit c) To fd, remove voltage sources from circuit by replacg them with shorts. Calculate the total resistance between the load termals. is equal to. [(Ω 6Ω) 0Ω] Ω. Ω 8 To fd V for the 0 volt dc source, remove the ac source by replacg it with a short. emove the load resistance and calculate the voltage that is present between the load termals. Fd the total resistance with only the 0 volt dc source the circuit. tot _0V _ dc [( 0Ω Ω) 6Ω] Ω 7. Ω 7 Fd the voltage between ohm and 0 ohm resistor. ( 0Ω Ω) 6Ω 0V 6.V 7.7Ω V _ 0 Fd the voltage across the ohm resistor. Ω V _ dc _ 6.V 5.55V Ω dc To fd V for the 0 volt ac source, remove the dc source by replacg it with a short. emove the load resistance and calculate the voltage that is present between the load termals. Fd the total resistance with only the 0 volt ac source the circuit. tot _ 0V _ ac [( 0Ω Ω) Ω] 6Ω 8. Ω 6 Fd the voltage between ohm and 0 ohm resistor. ( 0Ω Ω) Ω V _ 0 0V _ ac s 5t 6.V 8.6Ω Fd the voltage across the ohm resistor. Ω V _ ac _ 6.V 5.55V Ω ac The total Theven voltage is 5.55Vdc 5.55Vac s5t
11 The orton current is: 5.55Vdc 5.55V.8Ω ac s 5t _ V _ 5
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