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

• A. 5050 Ω
• B. 6050 Ω
• C. 4450 Ω
• D. 5550 Ω

Answer 1

Answer: (C)

4450 Ω

Answer 2

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.