Could someone please help me with the bottom part of this? The battery was 3.2 V

Question

Could someone please help me with the bottom part of this? The battery was 3.2 V, across a 22 kOhm series resistor and two resistors in parallel to each other, a 47 kOhm parallel resistor and a 100 kOhm parallel resistor. The three voltage drop measures at the top were taken from a voltmeter, and now I need to find a calculated value for V 1, 2, and 3 and I 1, 2, and 3.

Eatery 3.2v resst orss hve cmploted creut meuure the salucs f the potentiul drogs acrthe crcut measure the salues of the potent ual drops across each of the Ose you Series vin Volage drog across the 22 40) resistor Parniel v,-voltage drop across the 47 a resistir-lbav funtiel V:-voltage drop, across the 47 dresistir the crcut analysis Now, ssuming the value of voltage provided by the batteries, carry through the ercuit analysis to determine the ealeulated values for the Vi. V and Vi. Using Ohm's Law determine the calculated values fxr i., 1. and i (tfe curene denugh the respective resistances). Show your rkeie V 112 VI tcalculated) V2 (calculated) . V3 (calculated) Ii (Calculated)- 12 (Calculated)- 13 (Calculated)-

Explanation / Answer

here,

equivalent resistance , req = r1 + r2*r3/(r2+r3)

req = 22 + 47 * 100/147

req = 53.97 kohm

current through battery , I = V/Req

I = 0.059 mA

current through r1 is I1 = I = 0.059 A

current through r2, I2 = I * r3/(r2 + r3)

I2 = 0.04 mA

current through r3 , I3 = I * r2/(r2+ r3)

I3 = 0.019 mA

voltage drop across r1 , V1 = I1 * r1

V1 = 1.3

voltage drop across r2 , V2 = I2 * r2

V2 = 1.9 V

voltage drop across r3 , V3 = I3 * R3

V3 = 1.9 V

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