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GUJCET · Physics · Moving Charges and Magnetism
A galvanometer of resistance \(50 \Omega\) is connected to a battery of 8 V along with a resistance of \(3950 \Omega\) in series. A full scale deflection of 30 division is obtained in the galvano meter. In order to reduce this deflection to 15 division the resistance in series should be _________ \(\Omega\).
- A 7900
- B 1950
- C 2000
- D 7950
Answer & Solution
Correct Answer
(D) 7950
Step-by-step Solution
Detailed explanation
D
from \(I =\frac{ V }{ R + G } \alpha \phi\) (division)
\(\frac{\frac{V}{R+G}}{\frac{V}{R^{\prime}+G}}=\frac{\phi_1}{\phi_2}\)
\(\begin{aligned} \therefore \frac{R^{\prime}+G}{R+G} =\frac{30}{15} \\ \therefore \frac{R^{\prime}+50}{3950+50} =2 \\ \therefore R^{\prime}+50 =8000 \\ \therefore R^{\prime} =7950 \Omega\end{aligned}\)
from \(I =\frac{ V }{ R + G } \alpha \phi\) (division)
\(\frac{\frac{V}{R+G}}{\frac{V}{R^{\prime}+G}}=\frac{\phi_1}{\phi_2}\)
\(\begin{aligned} \therefore \frac{R^{\prime}+G}{R+G} =\frac{30}{15} \\ \therefore \frac{R^{\prime}+50}{3950+50} =2 \\ \therefore R^{\prime}+50 =8000 \\ \therefore R^{\prime} =7950 \Omega\end{aligned}\)
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