CBSE Class 10 Answered
![question image](http://images.topperlearning.com/topper/new-ate/089798ck5a7d8886e8f40.png)
Asked by piyushmalik024 | 18 May, 2019, 10:12: AM
i) The resistors R2 and R4 are in series and R3 is connected in parallel to these resistors.
Thus,
Rs = R2 + R4 = 15 + 15 = 30 Ω
![1 over R subscript p equals space 1 over R subscript s plus 1 over R subscript 3 space
fraction numerator begin display style 1 end style over denominator begin display style R subscript p end style end fraction equals 1 over 30 plus 1 over 20 space
T h u s comma space
R subscript p space equals space 12 space capital omega](https://images.topperlearning.com/topper/tinymce/cache/3685acd36e2f30d853c0547779d597a5.png)
Thus,
Effective resistance
Requivalent = 12 + 20 + 8 = 40 Ω
Thus, equivalent resistance of circuit = 40 Ω
iii) Total current through circuit,
V = I R
Total resistance = 40 Ω
Total potential difference = 12 V
Thus,
I = V/R
I = 12/40 = 0.3 A
Thus, total current through circuit = 0.3 A
iii) Current across each resistor:
The current across R3 = 20 Ω resistor is,
I = V/R = 3.6 / 20 =0.18 A
The current across R2 and R4 = 15 Ω resistor is,
I = 0.3 - 0.18 = 0.12 A
Current through R1=20 Ω and R5 = 8 Ω is 0.3 A.
iv) Voltage across each resistor,
The voltage drop across R1= 20 Ω,
V = 0.3 x 20 = 6 V
Voltage across R5 = 8 Ω resistors is,
V = 0.3 x 8 = 2.4 V
Voltage drop across combination of R2, R3 and R4
V = 0.3 x 12 = 3.6 V
Thus, voltage across R3 is 3.6 V.
Voltage aross R2 and R4 individually will be,
V = 0.12 x 15 = 1.8 V
Answered by Shiwani Sawant | 18 May, 2019, 12:52: PM
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