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The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -
  • a)
    1.0 A
  • b)
    0.2 A
  • c)
    0.1 A
  • d)
    2.0 A
Correct answer is option 'B'. Can you explain this answer?
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Calculation of Current Drawn from the Cell:
- The formula to calculate the current drawn from the cell in a Wheatstone's bridge is given by:
\[ I = \frac{E}{(r + R_{eq})} \]
Where,
- I is the current drawn from the cell
- E is the e.m.f. of the cell (7 V)
- r is the internal resistance of the cell (5 ohm)
- \( R_{eq} \) is the equivalent resistance of the bridge

Calculating Equivalent Resistance:
- The equivalent resistance of the Wheatstone's bridge can be calculated using the formula:
\[ R_{eq} = \frac{PR}{Q + R} \]
Where,
- P, Q, R, and S are the resistances in the arms of the bridge
- Substituting the given values:
\[ R_{eq} = \frac{(10)(90)}{30 + 30} = \frac{900}{60} = 15 ohm \]

Calculating Current Drawn:
- Substituting the values of E, r, and \( R_{eq} \) in the formula:
\[ I = \frac{7}{(5 + 15)} = \frac{7}{20} = 0.35 A \]
- However, since the galvanometer resistance is 50 ohm, the effective resistance in the circuit would be \( 15 + 50 = 65 ohm \)
- Recalculating the current drawn using the corrected resistance:
\[ I = \frac{7}{(5 + 65)} = \frac{7}{70} = 0.1 A \]
Therefore, the current drawn from the cell is 0.1 A, which corresponds to option C.
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The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be -a)1.0 Ab)0.2 Ac)0.1 Ad)2.0 ACorrect answer is option 'B'. Can you explain this answer?
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