Question

2:17 p. m. @ 132% Listo Online_Lab_on_Circuits_with_... ☺ Lab #5 : Circuits with Resistors and Capacitors Name: I. Simple circuits with resistors Imagine that you are given three resistors, each with a value R-300 Ohms and a DC power supply a battery) of 12 V. NOTE: In a normal lab you will be required not only to make the calculations but also to wire the circuits on a breadboard, measure the currents and compare your measurements with your predicted values of the currents. In the expraordinary quarantine situation associated with COVID-19 that part is dropped out. Just calculate the circuits, embed the pictures of the circuits in the spaces provided and upload your lab on Moodle in the designated Drop Box. 1. How would you connect these resistors if you would like to draw the largest current from the battery? What is the value of this current? Calculate the total power delivered by the battery to the circuit as well as the power dissipated to heat in each resistor. Draw a schematic of the circuit below and do the calculations to back upi your statements. 2. How would you connect these resistors if you would like to draw the smallest current from the battery? What is the value of this current? Calculate the total power delivered by the battery to the circuit as well as the power dissipated to heat in cach resistor. Draw a schematic of the circuit below and do the calculations to back upi your statements.

Answer 1

145 t. To doan connected in largest current the resistors have to be Series, parallel. R&R ER Current bon drawn from the source is, In Hot = 0·12 A. Total power delivered by the battery : Po = V. 1 = 12x0.12 W = 1044W. Power dissipated through. . Current through each 2. To doran min. should be man. be connected in - 0.04 x 300 W - 0.48 W. current, the equivalent resistance Hence the a resistors have to series. Tue 1 ŠR 12 3x300 3oo A Power ł 13.3 mA delivered by the battery Pga v. I = 12x 13.3 x 10-3 w. = 0.16 W. dissipated through each resistor. P = SRR = IR = (13.3x103) x 300 w ł 0.053 W. Power

The 3. laulu The schematic is as follows ! in Equivalent resistance ' I A3 94 Rega R+ 2 = 450.2 22 A 00027A Power delivered by the battery, Pp = I.V. = 0.027x12 w - 0.324W. Power dissipated by the upper resistor, P. = IR = 0.027 x 300 W = 0·22 W. Power dissipated by each of the resistor connected in parallel is. - p'= ( R = ( 00029 x 300 w 5 0055 w. 4. The schematic diagram will be as follows. Rex = QRIR = BRAR - 2x3000 a 3 x 300 - 200 a Ir 0 A = 0·06 A Power delivered by the battery. Pp 2 V. I 1 2 12 x 006 w Power dissipated the resistors in series is, Psa C) x Rate 12W 0·12 W. Power dissipated by the sea 4x300 wa Po - Ră W = 0.48W - 072W. " 6

FREE SEP const. sec of the circuit is, Ra R. Ca 5x10² x 12 x 10 9 sec = 60x10b sec. 260 vs.