Question:

A galvanometer of resistance 50 Ω is connected to the battery of 3 V along with a resistance 2950 Ω in series. A full scale deflections of 30 divisions is obtained in galvanometer. In order to reduce this deflections to 20 division, the resistance in series should be

Updated On: Apr 20, 2024
  • 5050 Ω
  • 6050 Ω
  • 4450 Ω
  • 5550 Ω
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The Correct Option is C

Solution and Explanation

Let's calculate the current for the full-scale deflection:
\(I_1 = \frac{V}{R_{\text{gal}} + R_{\text{series}}}\)

\(I_1 = \frac{3V}{50\Omega + 2950\Omega}\)

\(I_1 = \frac{3V}{3000\Omega}\)
\(I_1 = 0.001\, \text{A}\)
Now, let's calculate the current required for a 20 division deflection:
\(I_2 = \frac{2}{3} \cdot I_1\)

\(I_2 = \frac{2}{3} \times 0.001 \, \text{A}\)
\(I_2 = 0.000667 \, \text{A}\)
To find the resistance needed, we can use Ohm's law:
\(V = I \cdot R\)
For the 20 division deflection:
\(0.000667\, \text{A} \times (50\,\Omega + R) = 3\, \text{V}\)
Solving this equation for R, we get:
\(0.000667\, \text{A} \times R = 3\, \text{V} - 0.000667\, \text{A} \times 50\,\Omega\)

\(R = \frac{{3\, \text{V} - 0.000667\, \text{A} \times 50\,\Omega}}{{0.000667\, \text{A}}}\)
Calculating this expression gives:
\(R ≈ 4,449.25 Ω\)
Rounding this value, we get 4,450 Ω.

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