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- The diagram in the figure shows a simple circuit consisting of a battery and a load consisting of two resistors.The battery voltage is UB = 56 V and the value of the resistors is:R1 = 1800R2 = 5400a) The ammeters in the circuit show the following value:A1 shows valueA2 shows valueA3 shows valueb) In the load, the effect developsc) The total resistance of the load isIn the circuit shown below, if R1=5 ohm, R2=10 ohm and Vin = 2V. Then the current (I) will equal to: Vino Vout R1 R2 O a. -0.6A O b. 0.4A O c. -0.2A O d. -0.4A Oe. 0.6A150 100 + VA- a -4V AVA -7V 81. B b Figure 2 shows a simplified model of a gas-discharge lamp. One characteristic of these lamps is that they exhibit negative resistance; in other words, as current increases the voltage drops further, making such lamps inherently unstable. As such, a current-limiting ballast is required. For the connection shown in the figure: 1. Find the equivalent Thevenin circuit of the lamp. 2. Find the ballast resistance needed to limit the current drawn from a 24-volt source to 6 amperes.
- Estimate voltage E2 when E1=90sin2400t V, R1=R2=20 ohms, L1= L2=3mH and M=0.15mHTwo resistors R1 and R2 of 12 Q and 6 N are connected in parallel and this combination is connected in series with a 6.25 Q resistance R3 and a battery which has an internal resistance (in series with the battery e.m.f.) of 0.25 2. Determine the e.m.f. of the battery. (note: the current through 6 ohms is 0.8 A)INSTRUCTIONS: Solve the following problems. Show and COMPLETE solutions. Draw all CIRCUIT DIAGRAMS or the equivalent circuit (dummy circuit) as the case may be. Write your solutions on sheet/s of short bond paper. A battery is to consist of 20 identical cells. The emf of each cell is 1.5 V and the internal resistance is 0.20 ohm. This battery will be used to supply power to a 10 ohm lamp. Determine the current on the lamp if:1. The 20 cells are connected in series 2. The 20 cells are connected in parallel 3. The 20 cells are arranged 5 cells in series in 4 parallel rows.
- is made from a combination of A mixed resistors circuit as shown in Figure parallel and series circuits. The values of the resistances mentioned in the circuit are in Ohm (2) and the supply voltage is in Volt (V). Determine the: (i) Current through the circuit (I). (ii) Voltage drop (V) across the circuit. 0.6 0 4 0 0.5 04.-For-labeled currents, draw an arrow to show the direction of the actual positive current. For labeled voltages, circle the node that is at the actual highest potential. Note that the positive and negative signs of the voltage represent the polarity of the probes of a voltmeter connected to the device. The value shown is the reading on the voltmeter. (segment of wire) 2A -2A -5A 2.de 3A TA Cov E 5V 12V + 3V -10V -201In the figure the current in resistance 6 is i6 = 1.33 A and the resistances are R₁ = R₂ = R3 = 1.56 02, R4 = 14.7 Q, R5 = 7.74 Q2, and R6 = 3.81 Q. What is the emf of the ideal battery? 80 R₁ Number i www R₂ www R₂ Units www R₁ www R₂ i6 www R6
- The figure below shows three resistors (R- 13.5 0, R -7.95 0, and Ry- 12.5 0) and two batteries connected in a circuit. 40.0 V R 22.0 V R3 (a) What is the current in each of the resistors? A I2 = A A (b) How much power is delivered to each of the resistors? P1- P2- P3-In the circuit shown below, if R1=1 ohm, R2=6 ohm and Vin = 1V. Then the current (I) will equal to: Vin +1 Vout R1 R2 O a. 0.4A O b. -0.6A O c. -0.4A O d. 0.6A O e. -1AA 20-ohm resistor is connected in parallel with a variable resistor R. The parallel combination is then connected in series with a 4-ohm resistor and connected across a 240 V source. Determine the minimum value of Rif the power of R is equal to the power taken by the 4-ohm resistor.