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Use the circuit shown to answer the following questions:?
a) Write KCL at points B&F
At B:At F:b) Write KVL for the following Loops :Loop BGFB:Loop BCDGB:Loop AFGA:3. Solve the equations simultaneously to determine the real values for i1, i2, i3. Hint : no need to convert the resistance from K Ω to Ω . The answer directly will be in mA.I1=I2=I3=4. Determine the voltage across each resistance:V 3.3kΩ =V1.8kΩ =V5.6kΩ=( THE LAST 2 QUESTIONS ARE NOT IN THE PICTURE SO I WROTE THEM ABOVE, THANK U FOR HELPING)
1 Answer
- oldschoolLv 72 months ago
Loop analysis: Assume two cw currents i1 and i2
Loop1
-8.9i1 + 3.3i2 = 10 >>>>> i2 = (8.9i1+10)/3.3
Loop 2
3.3i1 - 5.1i2 = 6 - 10 = -4 Substitute i2 = (8.9i1+10)/3.3
3.3i1 - 5.1(8.9i1+10)/3.3 = -4
i1(3.3 - 5.1*8.9/3.3) = -4 + 51/3.3
i1 = (-4 + 51/3.3)/(3.3 - 5.1*8.9/3.3) = -1890/1725 ≈ -1096µA
i2 = 130/1725 ≈ 75.4 µA
i1-i2 = 2020/1725 ≈ 1171µA
V5.6 ≈ 6.14v
V3.3 ≈ 3.86v
V1.8 ≈ 0.136v
Compare power delivered to consumed:
10*2.02/1725 - 6*0.130/1725 = 11.258 mW
(1.89/1725)²*5.6 + (2.02/1725)²*3.3 + (0.13/1725)²*1.8 = 11.258 mW Checks
Node Analysis with just one node V at the top of the ckt
KCL at node V
V/5.6 + (V-10)/3.3 + (V-6)/1.8 = 0
V(1/5.6 + 1/3.3 + 1/1.8) = (10/3.3 + 6/1.8)
V = (10/3.3 + 6/1.8)/(1/5.6 + 1/3.3 + 1/1.8) = 10584/1725 ≈ 6.136
The voltage across all resistors ≈ 6.14
i5.6 = 6.136/5.6e3 ≈ 1096µA down
i3.3 = (6.136 - 10)/3.3e3 ≈ 1171µA up
i1.8 = (6.136 - 6)/1.8e3 ≈ 75.4µA down
Same answers.
