Question

FIGURE 10.1 LAB PREPARATION Review Section 6.3 of Electronics Technology Fundamentals MATERIALS 1 DMM 1 variable de power supply Pretoboard 9 resistors: 100 2247 K2, 10 kA (2) X2 XKA. 1 2 3 and 1009 1 100 potentiometer PROCEDURE 1. Calculate the no-load output voltage and bleeder current for the circuit shown in Figure 10.2. Enter your calculations in Table 10.1. SR 10kn 10 V 6R I 10 KO FIGURE 10.2

I am having trouble calculating the loaded current (IL) and the bleeding current (I2)

Answer 1

The main fast in this problem is " Measuring the vollage across the resistances with DMM " After measuring the voltage across the resistance, calculate current through resistance by formula I. V measured . Resistance valu. Step- the In the given No load voltage problem and you have alredy rdaulated bleekor curent (E). 2 In this he is asking , connect the load of look, Azk, lik and 100r measured the voltage across the load and find bleeder current (Is) and find load current es Scanned with fast CamScanner

Aphor maneeding is like below the land mesilanes, the circuit e TV 3 RL L NL for her all you the have alredy measured different values of load voltage load resistances. - Since Rq and Re voltage across the lood vollage vi are connected in bothe resistantes parallel, so the ax same as not? Ferin procedure point is, he clearly given that, measure load voltage f you already done it]. After that he is saying that by using Ÿ , calculate load current [Eland Sreeder current [7] so we know the voltages across the resistance rul so current through the resistance is calculated for can seregnet value in sto fable by 'T V I Scanned with RE

% change so Table 10.2 .. By using measured value of a ... I'm calculating It and I In this R = R = loker ; your are always RL N / Change IL look 4.75 - 4.18 = 0.047SMA your or alxdy calculata left that So if Iz 475 -0.475mA 4-75. Tot Tot 100k 100k 0-047SMA 0.4515 MA 0.451 Ark - 4.515 = 0.016 mA 4.515 4.515 40% 0.8328 - 08328 MA 0.8328 0.0054 Ik 100 0.8318 0:08287 - - 0.03287 0.01287 700 0.920 MA Vt 0.09207. 0-00910 OF MA Table 10.3 12 = R₂ = lkn Ig Be į % change 4.1963 – 4-963 MA 4.903 700k 0-04963 MA 4.963 look Ik - 4.935 = 0.105 mA = 4.935mA 4tk 4.937 471 3.372 3.373 MA 3.373 3373 = 3-373 MA 1 Ik 100 0.8999 - 8.799 MA 0-879 Too 0.8791 = 0.8791MA Scanned with *** CamScanner