can anyone help with this?
As we can see from table
NO resistors R2 and R4 dont have same voltage or current
if we see the circuit with out ammeter or voltmeter we can tell resistors R2 is in parallel connection with resistors R3 and R4 which are in series there fore voltage across R2 will be same as sum of voltage of R3 and R4 not R4 only. Current in R2 and R4 will be equal if resistor value of R2 is equal to sum of resistor value of R3 and R4 here R2 is 2k ohms and R3+R4 is 1.51K ohms which are not equal, there fore their current are not equal.
Consider a circuit with four resistors in series. R1 R2 R3 R4 The battery has a voltage of 12 V. The resistor values are R1-800 Ω. R2-600 Ω R3-500 Q . R4-500Q Complete the following table by noting the voltage drop across each listed resistor. Resistor (Q) Voltage Drop (V) R1 00Select R3500Select ] 00 Select ] R2 R4
Consider the circuit below. R1 = 3 ohms, R2 = 6 ohms, R3 = 2 ohms, R4 = 4 ohms, and the battery has a voltage of 12 V. Answer the following Pick the best answer for each True/False answer 1. R1 and R2 will both have a voltage across each of 12V. True or False. 2. R1 and R2 will both have the same current traveling through them. True or False. 3. R3 and R4 will both have a...
1. Find the voltage across resistor R3 R4 Sko V1 R3 - 12 $1. 00 2. Find the current through and voltage across R6 R1 R2 1.sk = V2 12V 1.00 R4 R8 4700 1.Ok R7 $1.0ko R6 2.25kn 3. Find the voltage across and current through R3 R1 R2 4.7k 1.Ok 12V 1.5k) R5 W 4.7kQ R4 WW 1.5k R6 4. Find the voltage across R2 and the current through R6 R7 R5 R3 1.0k 1.0kg vi 12V R6...
Suppose R1=4Ω, R2=2Ω, R3=8Ω, R4=10Ω, 5=6Ω, R6=16Ω, E1=9V and E2=11V. Part A: Use Kirchhoff's voltage law to find an equation for the voltage drop around loop 3 Part B: Find the loop currents i1i1, i2i2 and i3i3 (positive for clockwise direction) Part C: Which direction does current flow across each of the resistors? Match with the following choices. Possible choices : A:from left to right B: from right to left C: no current D: from top to bottom E: from...
R1 R3 R2 R4 W The circuit in the diagram contains one battery and four resistors, labeled 1 through 4. ?Which of the following is true about the way that the circuit elements are arranged Resistors 1 and 2 are in parallel while resistors 3 and 4 are in series Resistors 1 and 4 are in parallel while resistors 2 and 3 are in series Resistors 1 and 4 are in series, while resistors 2 and 3 are in parallel...
R1 1.5K R3 3K V1 50V R2 6k R4 6K R1 Total R2 R3 R4 Current Voltage Resistance 1.5K Power NA 6K 3K 6K
R1 VB1 VB2 5R5 R2 R4 In the circuit above, the current through R5 (labeled as I5) is 15 mA. Given this information, find the magnitude of the current through the other resistors and the magnitude of the voltage drop across the second battery VB2 VB1 = 3 volts I5 15 mA R1 160Ω R2 = 300Ω R3 = 450Ω R4 270 R4 VB2
5. Consider the next circuit with R1 1 kQ, R2 2.2 kQ, R3 3.3 kQ, R4 4.7 kQ and E 10V Using the Ohm's law, determine the currents through each of the four resistors and record the values in the next table under the Theory column. Note that the larger the resistor, the smaller the current should be. Also determine and record the total supplied current and the current IX Note that this current should equal the sum of the...
E: 10 R1: 2.2k ohm R2: 4.6k ohm R3: 3.3k ohm R4: 6.8k ohm Questions to Answer: 01: Choose the appropriate resistors and voltage sources and find the thevenin voltage and thevenin resistance both theoretically and practically and verify your results. Consider Raas the load resistance. Demonstrate all the working b. a. Fill the tables with appropriate measured and calculate values. Draw the thevenized circuit c. Original Circuit: R Load) VLoad Theory Vond Experimental Deviation R1 R3 R2 R4 Thevenized...
Problem 3: Determine explicitly the voltage across R3 in the circuit below R2 R1 R3 R4 (Hint: R2 and R3 are in parallel with R1... this combination is in series with R4. The total resistance can give you the total voltage drop AND then you can just analyze as a voltage divider.)