A galvanometer has a 50 division scale. The b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I$ be the full scale deflection current and $V = 2\,V$ be the battery voltage.In case $1$,when $R_1 = 2400\,\Omega$,deflection $	heta_1 = 40

Vedclass · Accepted Answer

Current sensitivity of galvanometer is $20\,\mu A/	ext{division}$.

Vedclass · Answer

(D) Let $I$ be the full scale deflection current and $V = 2\,V$ be the battery voltage.In case $1$,when $R_1 = 2400\,\Omega$,deflection $	heta_1 = 40$ divisions.$\frac{40}{50} I = \frac{V}{G + R_1} \Rightarrow \frac{4}{5} I = \frac{2}{G + 2400} \dots (1)$In case $2$,when $R_2 = 4900\,\Omega$,deflection $	heta_2 = 20$ divisions.$\frac{20}{50} I = \frac{V}{G + R_2} \Rightarrow \frac{2}{5} I = \frac{2}{G + 4900} \dots (2)$Dividing $(1)$ by $(2)$:$\frac{4/5 I}{2/5 I} = \frac{G + 4900}{G + 2400} \Rightarrow 2 = \frac{G + 4900}{G + 2400}$$2G + 4800 = G + 4900 \Rightarrow G = 100\,\Omega$.Substituting $G = 100\,\Omega$ in $(1)$:$\frac{4}{5} I = \frac{2}{100 + 2400} = \frac{2}{2500} = \frac{1}{1250}$$I = \frac{5}{4} 	imes \frac{1}{1250} = \frac{1}{1000}\,A = 1\,mA$.Current sensitivity $= \frac{I}{50} = \frac{1\,mA}{50} = 0.02\,mA/	ext{division} = 20\,\mu A/	ext{division}$.For a deflection of $10$ divisions:$\frac{10}{50} I = \frac{V}{G + R} \Rightarrow \frac{1}{5} 	imes 10^{-3} = \frac{2}{100 + R}$$100 + R = 10000 \Rightarrow R = 9900\,\Omega$.Thus,option $D$ is correct.

$A$ galvanometer has a $50$ division scale. The battery has no internal resistance. It is found that there is a deflection of $40$ divisions when $R = 2400\,\Omega$. The deflection becomes $20$ divisions when the resistance taken from the resistance box is $4900\,\Omega$. Then we can conclude:

Similar Questions

$A$ moving coil galvanometer of resistance $100 \Omega$ is used as an ammeter using a shunt resistance $0.1 \Omega$. The maximum deflection current in the galvanometer is $100 \mu A$. Find the total current in the circuit so that the ammeter shows maximum deflection. (in $mA$)

$A$ galvanometer can be converted into an ammeter by connecting:

$A$ galvanometer of resistance $40 \Omega$ gives a deflection of $10 \text{ divisions per } mA$. There are $50 \text{ divisions}$ on the scale. The maximum current that can pass through the circuit when a shunt resistance of $2 \Omega$ is connected is: (in $\text{ mA}$)

Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module