Download 1 ENGR 120 - Possible exam or example circuit problems Stan

1 ENGR 120 ‐ Possible exam or example circuit problems Stan Cronk 1. The Boe‐Bot kit includes resistors with values of 220Ω, 470 Ω, 1K Ω, 4.7K Ω, AND 10K Ω. All the resistors are rated to dissipate up to ¼ ‐Watt of power. The Boe‐Bot can take power from voltage sources ranging from 6V to 12V. If you are using a 12V power source, and you want to connect a single resistor directly from Vin to Vss, what is the minimum value of resistor in the Boe‐Bot kit that you can use that would NOT break? R = V2 / P = 122 V2/ 0.25W = 576 Ω. Must use at least 1k Ω resistor. 2. You have designed a circuit that uses a 25‐Ω resistor; unfortunately, the resistor is broken. Fortunately, your professor has given you a package of 50 100‐Ω resistors. Design a resistor network using only resistors your professor has provided to replace the 25‐Ω resistor. Req = 100 Ω / n for n resistors in parallel 25 = 100 Ω / n n = 100 Ω / 25 = 4 resistors in parallel 3. You have designed a circuit that uses a 220‐Ω resistor; unfortunately, the resistor is broken. Fortunately, your professor has given you a package of 75 100‐Ω resistors. Design a resistor network using only resistors your professor has provided to replace the 220‐Ω resistor. Note that for resistors of the same value, Req_series = R*n and Req_parallel = R/n Use 2 resistors in series (200 Ω) + 5 resistors in parallel (20 Ω) 4. Referring to the circuit in Figure 1: (a) Find the total resistance in the circuit. (d) Find the current through R2. (b) Find the total current produced by the voltage source. (e) Find the current through R3. (f) Find the current through R4. (c) Find the current through R1. (Figure 1) 2 (a) Req = 16 ohms (R1 is in parallel with R234, the sum of resistors 2, 3 and 4) (b) IT = 3.125 A (50 V / Req) (c) IR1 = 2.5 A (50V / R1) (d) IR2 = .625 A (calculate either by subtracting IR1 from IT or by dividing 50 V by R234) (e) IR3 = .625 A (same current flows through all resistors in series, so IR3 = IR2) (f) IR4 = .625 A (same current flows through all resistors in series, so IR4 = IR2) 5. Referring to the circuit in Figure 2, if I1 = 1 A, I2 = 2 A, I3 = 5 A: (a) Find the value of resistance of R1. (d) Find the total current. (b) Find the value of resistance of R2. (e) Find the power generated by the voltage source. (c) Find the value of resistance of R3. (Figure 2) (a) R1 = 20 ohms (20 V / 1 A ) (b) R2 = 10 ohms (20 V / 2 A) (c) R3 = 4 ohms (20 V / 5 A) (d) IT = 8 A (sum of I1, I2 and I3) (e) PS = 160 W (20 V * 8 A) 6. The total power generated by the voltage source should be equal to the power dissipated by all the resistors, whether they are in series, parallel, or a combination of series and parallel. Verify that this is true by referring to the circuits in Figures 3a and 3b: 3 (a) Find the total equivalent resistance of each circuit. (b) Find the total current coming out of the voltage source in each circuit. (c) Find the total power generated by the voltage source in each circuit. (d) Find the power dissipated by each resistor in each circuit. (e) Does the total power generated by the voltage source in each circuit = the total power dissipated by the resistors in each circuit? (Figure 3: Circuit S) (Figure 3: Circuit P) (a) Req (circuit S) = 20 ohms; Req (circuit P ) = 3.75 ohms (b) IT (circuit S) = 5 A (100 V / Req) ; IT (circuit P) = 26.667 A (100 V / Req) (c) PS (circuit S) = 500 W; PS (circuit P) = 2,666.7 W (P = IV or I2R) (d) Circuit S: PR1 = 375 W (I2R); PR2 = 125 W (I2R) Circuit P: PR1 = 666.67 W (V2/R); PR2 = 2000 W (V2/R) (e) In each case, the sum of the powers dissipated or consumed by the resistors does indeed = the power generated by the voltage source. 4 7. Referring to Figure 4, (a) Find the total resistance in the circuit (b) Find the total current generated by the voltage source (c) Find the current in R3 (d) Find the current in R4 (e) Find the power dissipated by R4. (Figure 4) (a) Req = 28.28 ohms (R12 = 12 ohms; R123 = 52 ohms; R56 = 12 ohms; R456 = 62 ohms) (b) IT = 3.536 A (100 V / Req) (c) IR3 = 1.923 A (easiest way to calculate – combine R1, R2, and R3 into a single resistor (R123); then the current through R3 = the current through that branch, which will be 100 V / R123) (d) IR4 = 1.613 A (IT – IR3) (e) PR4 = 130.088 W (IR42Rr) 8. Fill in the blank. (a) All resistors in series have the same current. (b) According to Kirtchoff’s Voltage Law, the sum of all voltages around a closed loop equals zero. (c) If a resistor dissipates 10 W with 1A of current flowing through the resistor, the value of resistance must be 10 ohms and the voltage across the resistor must be 10V. (d) If two resistors (R1 and R2) are in parallel, and R1 has a higher value of resistance than R2, then the current through R1 will be _________________ (higher than, lower than, or the same as) the current through R2.