Calculate the following from the provided circuit diagram. a) Total resultant resistance b) The total current c) Voltage across 3.6 Please give the correct answers as ,it's really important. Don't answer it with something sillyanswers just for the points I provide.
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Answers
Answer:
a) The resultant resistance is 9Ω.
b) Total current in the circuit is 0.5 A.
c) Voltage across 3.6 Ω resistor is 1.8 V.
Explanation:
a) Total resultant resistance :
The resistor R2 is parallel with R4 and R5. R4 and R5 are in series.
⇒ R series = R4 + R5
⇒ R series = 1Ω + 3Ω
⇒ R series = 4Ω
⇒ R4,5 = 4Ω
The resistor R2 is parallel with R4,5
⇒ 1/RP = 1/R2 + 1/R4,5
⇒ 1/RP = 1/6 + 1/4
⇒ 1/RP = (4 + 6)/24
⇒ 1/RP = 10/24
⇒ RP = 2.4 Ω
R1, RP and R3 are in series,
⇒ 3 + 2.4 + 3.6
⇒ 3 + 6
⇒ 9 Ω
Therefore, the resultant resistance is 9Ω.
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b) Total Current in circuit :
⇒ I = V/R
⇒ I = 4.5/9
⇒ I = 0.5 A
Therefore, total current in the circuit is 0.5 A.
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c) Voltage across 3.6 Ω :
The resistor of 3.6 Ω is in series. Which means the current in the circuit is constant.
⇒ V = I × R
⇒ V = 0.5 × 3.6
⇒ V = 1.8 V
Therefore voltage across 3.6 Ω resistor is 1.8 V.
Required Answer :-
a) Total resultant resistance
Since, R4 and R5 are in series. The equivalent resistance of series is given by
Rn = R4 + R5
Rn = 1 + 3
Rn = 4 Ω
Now, R2 is connected with the R4 and R5 in parallel. The equivalent resistance for parallel is given by,
1/Rn = 1/R2 + 1/R4 and R5
1/Rn = 1/6 + 1/4
1/Rn = 4 + 6/24
1/Rn = 10/24
1/Rn = 5/12 Ω
Rn = 12/5
Rn = 2.4 Ω
Now
R1, R2 and R3 are in series
3 + 2.4 + 3.6
3 + 6
9 Ω
b) The total current
We know that
V = IR
4.5 = I(9)
4.5 = 9I
4.5/9 = I
1/2 = I
0.5 = I
Hence
Current is 0.5 Ampere
c) Voltage across 3.6
We know that
V = IR
V = 0.5 × 3.6
V = 5/10 × 36/10
V = 1.8 Volt