OHM'S LAW?
2016-02-29 8:20 am
3 resistors are connected in series to a battery. From left to right, the resistances are R1, R2 = 5 ohms, R3. The voltsge across R1 and R2 is 8V while the voltage across R2 and R3 is 4V. The equivalent resistance of the 3 resistors is 22 ohms. Determine R1, R3 and battery voltage. THANKS IN ADVANCE!!!
回答 (2)
2016-02-29 8:55 am
✔ 最佳答案
Ohm's law : V = I RI : current in the series circuit.
The equivalent resistance (in Ω) of the three resistors :
R₁ + 5 + R₃ = 22
R₁ + R₃ = 17 ...... [1]
Voltage (in V) across R₁ and R₂ :
I (R₁ + 5) = 8 ...... [2]
Voltage (in V) across R₂ and R₃ :
I (5 + R₃) = 4 ...... [3]
[2]/[3] :
(R₁ + 5) / (5 + R₃) = 2
R₁ + 5 = 2 (5 + R₃)
R₁ + 5 = 10 + 2R₃
R₁ - 2R₃ = 5 ...... [4]
[1] - [4] :
3R₃ = 12
R₃ = 4 (Ω)
Put R₃ = 4 (Ω) into [1] :
R₁ + 4 = 17
R₁ = 13 (Ω)
Put R₃ = 4 (Ω) into [3] :
I (5 + 4) = 4
I = (4/9) A
Battery voltage
= I R
= (4/9) × 22
= 9.8 V
Ans:
R₁ = 13 Ω
R₃ = 4 Ω
Battery voltage = 9.8 V
2016-02-29 8:43 am
Assigning i as current flow through circuit,
iR1 + iR2 = 8V
R1 + 5 = 8/i
R3 + 5 = 4/i
R1 + 5 + R3 = 22
R1 + R3 = 17
R1 = 17 - R3
17 - R3 + 5 = 8/i
-R3 + 22 = 8/i
R3 + 5 = 4/i
---------------------
27 = 4/i
i = 8/27 = 0.30 A
R1 + 5 = 8/0.3
R1 = 26.7 - 5 = 21.7 ohm
R3 + 5 = 4/0.3
R3 = 13.3 - 5 = 8.3 ohm
iR1 + iR2 = 8V
R1 + 5 = 8/i
R3 + 5 = 4/i
R1 + 5 + R3 = 22
R1 + R3 = 17
R1 = 17 - R3
17 - R3 + 5 = 8/i
-R3 + 22 = 8/i
R3 + 5 = 4/i
---------------------
27 = 4/i
i = 8/27 = 0.30 A
R1 + 5 = 8/0.3
R1 = 26.7 - 5 = 21.7 ohm
R3 + 5 = 4/0.3
R3 = 13.3 - 5 = 8.3 ohm
收錄日期: 2021-04-18 14:31:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160229002005AAHABNr