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CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 7 of 26
Marks: +1, -0
A 10 V10\ \mathrm{V} cell of negligible internal resistance is connected in parallel across a battery of emf 200  V\ \mathrm{V} and internal resistance 38 Ω38\ \Omega as shown in the figure. Find the value of current in the circuit.
OR
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm350\ \mathrm{cm}. When a resistance of 9 Ω\Omega is used in the external circuit of the cell, the balance point shifts to 300 cm300\ \mathrm{cm}. Determine the internal resistance of the cell.
Solution:  
Writing the equation
Finding the current
By Kirchoff's law, we have, for the loop ABDC,
200−38i−10=0200-38i-10=0
∴    i=19038A=5A\therefore\;\; i = \frac{190}{38} \mathrm{A} = 5 \mathrm{A}
OR
Stating the formula
Calculating rr
We have, r=(  l1l2−1)R=(  l1−l2l2)Rr = \left( \; \frac{l_1}{l_2} - 1 \right) R = \left( \; \frac{l_1 - l_2}{l_2} \right) R
∴    r  =(  350−300300)×9Ω\therefore\;\; r \; = \left( \; \frac{350-300}{300} \right) \times 9 \Omega
  =  50300×9Ω=1.5Ω\; = \; \frac{50}{300} \times 9 \Omega = 1.5 \Omega
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