CBSE Class 10 Answered
Physics
Asked by Sanchit | 18 Jun, 2019, 08:00: PM
Expert Answer
If we analyse the resistance network across the points A and B ( right side of AB ),
series combination of resistors R2 and R4 is parallel with resistor R3 .
Series combination of resistors R2 and R4 gives equivalent resistance 30Ω . Hence if this 30Ω is parallel to R3 of value 20Ω,
then we get equivalent resistance between A and B is, (30×20)/(30+20) = 12 Ω.
Modified circuit is shown in fig.2 by replacing the equivalent resistance across A and B.
Hence total resistance connected across 12V battery is 20Ω+12Ω+8Ω = 40Ω.
(as given above, 40Ω is the answer for question(1) i.e. equivalenet resistance of entire circuit )
Hence total current drawn from battery = 12V/40Ω = 0.3 A
It is shown in fig.1, main current i is divided as i1 and i2 .
Since equivalent resistance of R2 , R3 and R4 is 12Ω ,
potential difference VAB across A and B is given by, VAB = 12Ω × 0.3A = 3.6V
hence i1 × R3 = 3.6 V or i1 = 3.6/20 = 0.18A
since i = i1 + i2 , i2 = i - i1 = 0.3 - 0.18 = 0.12A
Current through resistance R1 = 0.3 A , voltage drop across R1 = 20 Ω × 0.3 A = 6 V
Current through resistance R2 = 0.12 A , voltage drop across R2 = 15 Ω × 0.12 A = 1.8 V
Current through resistance R4 = 0.12 A , voltage drop across R4 = 15 Ω × 0.12 A = 1.8 V
Current through resistance R3 = 0.18 A , voltage drop across R3 = 20 Ω × 0.18 A = 3.6 V
Current through resistance R5 = 0.3 A , voltage drop across R5 = 8 Ω × 0.3 A = 2.4 V
Answered by Thiyagarajan K | 18 Jun, 2019, 09:46: PM
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