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Two equal resistances R\(_{1}\) = R\(_{2}\)= R are connected with a 30 \(\Omega\) resistor and a battery of terminal voltage V. The currents in the two branches are 2.25 A and 1.5 A as shown in the figure. Then, what will happen? Please explain the answer in brief.
R\(_{2}\)= 15\(\Omega\)
R\(_{2}\) = 30\(\Omega\)
V = 36 V
V = 180 V
Posted by: Srivastava on 07.10.2017
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The correct answer is given by option (D).
Explanation: The potential drop across R2 and 30 Ohm will be same as they are in parallel. Now current across R2 is clearly 2.25-1.5 = 0.75 A. Thus,
0.75 x R2 = 30 x 1.5
Therefore, R2 = 60 Ohm = R1.
The equivalent resistance of R2 and 30 Ohm will be
R3 = (60 x 30)/(60 + 30) = 20 Ohm.
Therefore, the total resistance of the circuit is
R4 = R1 + R3 = 80 Ohm
Therefore,
V = 80 x 2.25 = 180 V
Answer A. R2 = 15 Ohm
Total Current is 2.25 A.
The current through the 30 ohm resister is 1.5 A.
Hence Current through the R2 should be equal to 0.75 A (2.25-1.5 = 0.75 A)
If R2 is taken as Unknown, Then R2 is given by,
By Current didsion method, 1.5 = 2.25 x 30 / (30+R2)
1.5 x (30+R2) = 67.5
45 + 1.5 R2 = 67.5
1.5 R2 = 67.5 - 45
R2 = 22.5 / 1.5
R2 = 15 Ohm
Since, total current in the circuit is constant, we get 2.25R = 0.75R + 1.5*30 .
Or, 1.5R = 1.5*30 .
Hence, R = 30 ohm Ans. (option B)
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