ELECTRICITY FOR TRADES (LOOSELEAF)
3rd Edition
ISBN: 9781260437454
Author: Petruzella
Publisher: MCG
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The voltage divider shown in the figure has a potential difference (V) between point A and B and a potential difference
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1. Design the voltage divider of figure 1 such that V, = V, if I = 5mA and E=72 V.
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- Use the general voltage divider formula to calculate the values of voltage drop for the following series-connected resistors. Assume a source voltage of 120 V. R1 = 1K Ω R2 = 2.2K Ω R3 = 1.8K Ω R4 = 1.5K Ω E1 =V E2 =V E3 =V E4 =Varrow_forwardFigure 5 shows a current series feedback amplifier. The sampled current signal is the output current, I, flowing in resistor, Rg, where it develops a feedback signal voltage, V across the resistor, Rg. Determine: a) the feedback gain, B = I. 1. Io the gain without feedback, A = Vs b) A c) the gain with feedback, Af 1+BAarrow_forwardQuestion 5 A certain circuit element has the current and voltage as follows: i = 10e-5000t A v = 50(1 – e-5000t) V Find the total energy transferred during t 2 Os.arrow_forward
- Using voltage divider for the below circuit Find v₂ ? U₁ 10 V + 01 R₁ 18 ΚΩ O R₂ 2 ΚΩ +arrow_forward3. Determine the total resistance and the total current. 100V 1 + 1002 M 15.2. M 1502 500 1052 1022arrow_forwardFind the current in and the potential difference across each element in the following circuit. 5.0 N 2.5 N 10 Ω ww ww ww 10 V 10 Ω 15 N 5.0 Ω -ww wwarrow_forward
- Calculate the output voltage VA when the voltage divider is loaded with R1= 7 KO VA=.... R1 1k2 U = 20 V R2 1 kΩ RL 7k 2 VA +arrow_forwardFind the Norton's equivalent current. 4 A O O O 5A 2A 1A 4A ↑ 10 Q2 WW 10 Q2 M Ww -20 22 O a barrow_forwardDetermine the total power dissipated in two resistors of the circuit. 1A († 500Ω A 31.25 W B 7.82 W C 39 W 100V + None of these I 5ΩΣ VR +5VRarrow_forward
- 4. Determine: Branch currents and power in each resistor using Kirchhoff's Laws R1 = 10 Q R3 = 30 Q R2 20 V A 20 Ω R4 = 40 Q R5 = 50 Q R6 = 60 Q R7 = 70 Q C 30 V 40 V R8 = 80 Q R9 = 90 Qarrow_forward(5) Use KVL and KCL simultaneously to find Ia in Figure 5a, 5b and 5c. 12 V 6 V 1 k2 2 k2 6 V 20 202 1 k2 2 kN 1 kN 4 V (a) Given I. = 2 mA (b)arrow_forwardQ VEL Ve → IR₂ R₂ Tea VBE DIMR HA Re Figure-1 1) Figure 1 shows a voltage divider circuit. By referring to figure - 1, Produce desgin equation for each resister (R₁, R₂, Rc and Re) based on the following information: VCE = 1/2 Vec, 1R2 = 10 18a, Ica = 2mA, Vcc = 12 V, VBFCON) = 0.65 V, V₁ = 0.1 Vcc • V₁ = V₁B + V₂ = 6+1·2= 7.2 V • Rc= V₂c-Vc/Ica = 12-7-2/2 = 2.4 ks2 take ß= loo, 1BQ = 1cQ/B = 2/100 0.02mA LEQ = IBA + Ico= 0·02+2=2.0 m A R₁ = V₁ / 1₁ = 1.2/2.02 = 0.59KR • VB = VBE + VE = 0.65 + 1.2 = 1.85 V •IR₂ = 10x 0.02 20.2m A • R₂ = VB/1R₂ = 1.85/0·2=9.25k By voltage division Rule, V₁ = Vcc R₂/R₁+R₂ R₁ + 9.25= 60 R. So.75 KQ lok s + VCE RE Figure-2 LKJ (A) RE=0.59K R₁ = 2.4K 22 2) Construct the circuit shown in figure 2, determine the value of Boc at the given Q-point. Vcc= 12V 2N3904 IMD lokn (2 FFL VCE=(V B IB + V66=0.6515 2 No. ● KUL in loop (2) |-6-10-³ x 1₂ x 10³ +12=0 I₁ = 6 A Boc Tc/10 BAC = 6/11-2376 DE Bpc = 0.533392 DC voltage, Vec - 12V collector to emitter voltage, VcR…arrow_forward
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