In the circuit of Fig. 4_2, let 1₁-6 A, 12-2 A, V- 5 V, R₁-5 02, and R₂-10 02. Which of the follow statements is true? 1₁ V₁ R₁ V₁ V₂ ww 1, R₂ 12
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- 1/1 By using the experiment of series connection: if the R1=50ohm, R2=100ohm and R3=150ohm.. the voltage drop at R2 is greater than R3 * and less than R1 true O false OEnter in the Voltage response as a numerical answer in a+bj format if R1=1122, R2=802, X1=9N, X2=4N, IS=(3+0j)A. V = Preview V A sample answer would be: 1+2j The relative tolerance for this problem is 10%. 50 R₁ Get help: Written Example jX₁ Ht P₂ jX2(a) Use Ohm's Law to derive an expression relating Vout to Vin using the values of the two resistors. For this calculation, assume that you have no load attached (i.e. RL → xx). Make sure to explain in words each step you take. Hint: Vout is just the voltage drop across R2. You can draw in a battery with AV = Vin if that helps you set up the relationships. R1 Vin R2 Vout (O Vin- Your answer should be in the form, Vout The term in parentheses is called the gain of the divider, or the divider ratio. in· ww
- + ww R1 Ω Rm i Click here for image B M R2 Ω Vs V + G ww R3 2 In the circuit above, R1 = 8.00, R2 = 10.00, R3 = 7.00, Vs = 2.00, and Rm = 6.002. I have ampere units. Find the voltage at node B with the respect to the reference ground node specified. Express your answer in volts and round it to two decimal places. Your answer must be purely numerical! Do not type any letters in the answer box.Compute for the value of VCEQ in the given circuit below: Given that: VBB = 4.0V Rв 3D 11.04k0 Rc = 2560 BDc = 100 Vcc =12V Rg Poc= 100 VBB Note: Express your answer in 3 decimal places, no need to include the unit. Example: If your calculated answer is 16.8395mA you should answer 0.017.Q3) For the network shown in the figure below, determine the following: a) re b) Zini and Zinz c) Zoj and Zo2 d) Av1, Av2, and AvT +20 V 6.8 ka 30 kQ 6.8 ka 30 ka 0.5 pF 0.5 uF 1150 B-150 1.5 ka 50 uF 1.5 ka 50 uF
- Now let's try using a pack of AA batteries (1.2 V output voltage and 0.2 Ohm internal resistance each) to replace the lead-acid battery. In order to match the car lead-acid battery, first, 10 such AA batteries need to be connected in series to form a 12V cell, while 50 such cells need to be connected in parallel (as shown in picture above; the total of 500 AA batteries are used). Find the power that is supplied to the same "car starter" using this pack of AA batteries. (Hint: for the second question, you need to find the total voltage and the internal resistance of the whole pack; you also need to find the resistance of the "car starter"; this can be done using the data of the first part of this problem)https://docs.goog X p tabs together, right-click a tab z5bCAq361TACoM7x7Q/formResponse R= 3 Ohm Vs- 12 v O Got it Remind me later Q6: For the following circuit, find the total resistance between points A,B (Reg) and the total current from point B to A if the applied voltage VAB equal 30 v ww 152 22 102 G F 20 2 10 2 40 S2 30 2 ww Reg 6 Ohm It= 5 AO Reg =4.5 Ohm It= 6.66 A Reg 3 Ohm It= 10 A Reg -6 Ohm It= -5 AO none of the mentioned ABE wwwA voltmeter having a f.s.d. of 100V and a sensitivity of 1.6 kohm/V is used to measure voltage V1 in the circuit of given Figure. Determine the value of voltage V1 with the voltmeter not connected, and the voltage indicated by the voltmeter when connected between A and B Α 40 ΚΩ Β 60 ΚΩ V₁ 100 V
- 4. For the following circuit, derive an equation for the voltage out (Vout) compared to the voltage in (Vin) as a function of R1 and R2. Show your steps. Hint: apply Ohm's law to both Vout and Vin. Vin R1 RZ VoutP/ calculate the effective resisfance of the following combination of vesistance between pomts A and B then find veltage drop across each resistance when BD of 6ov is appliad between Puints A and B AConsidering the circuit in the following figure, if V1 varies between -20 V and 0 V and Vb=5V, then the current through V1 is (assume ideal diodes): * D1 D2 4 ko 2 kO V1 Vo Vb -8 V O a. ((V1 +5)/4) mA for V1 2 -5 V and Zero for V1 s -5 V O b. ((V1 -5)) mA for V1 s - 5 V and Zero for V1 2 -5 V Oc. ((V1 +5)/4) mA for V1 s - 5 V and Zero for V1 2 - 5 V O d. ((V1 +5)/4) mA for V1 s 5 V and Zero for V1 2 5 V