Q.37. A voltmeter with internal resistance of can be used to measure upto 20 V. In order to increas

Q: What is the correct answer for JEE Main Physics Q37?

The correct answer is A. Sol. A voltmeter is a galvanometer in series with a high resistance. Let $I_g$ be the full-scale deflection current. For the initial range $V = 20 \text{ V}$, $V = I_g (R_g + R_s)$, where $R_g = x$ is the internal resistance of the voltmeter (which is the galvanometer resistance plus any series resistance already present). So, $20 = I_g (x)$. To increase the range to $V' = 30 \text{ V}$, we need t

Q: Q.37. A voltmeter with internal resistance of can be used to measure upto 20 V. In order to increase its measuring range to 30 V, the required modification is to ________

The correct answer is A. Sol. A voltmeter is a galvanometer in series with a high resistance. Let $I_g$ be the full-scale deflection current. For the initial range $V = 20 \text{ V}$, $V = I_g (R_g + R_s)$, where $R_g = x$ is the internal resistance of the voltmeter (which is the galvanometer resistance plus any series resistance already present). So, $20 = I_g (x)$. To increase the range to $V' = 30 \text{ V}$, we need t

Q: What are the options for this JEE Main Physics question?

A. connect resistor of $\frac{x}{2}$ in series with voltmeter. · B. connect resistor of $\frac{x}{2}$ in parallel to voltmeter. · C. connect resistor of $x \ \Omega$ in series with voltmeter. · D. connect resistor of $2x \ \Omega$ in parallel to voltmeter.

Question

Q.37. A voltmeter with internal resistance of  can be used to measure upto 20 V. In order to increase its measuring range to 30 V, the required modification is to ________

FirstInTest · Accepted Answer

Sol. A voltmeter is a galvanometer in series with a high resistance. Let $I_g$ be the full-scale deflection current. For the initial range $V = 20 	ext{ V}$, $V = I_g (R_g + R_s)$, where $R_g = x$ is the internal resistance of the voltmeter (which is the galvanometer resistance plus any series resistance already present). So, $20 = I_g (x)$. To increase the range to $V' = 30 	ext{ V}$, we need to add an additional series resistance $R'$. So, $30 = I_g (x + R')$. Dividing the two equations: $\frac{30}{20} = \frac{I_g (x + R')}{I_g x} \Rightarrow \frac{3}{2} = \frac{x + R'}{x} \Rightarrow 3x = 2x + 2R' \Rightarrow x = 2R' \Rightarrow R' = \frac{x}{2}$. Thus, a resistor of $\frac{x}{2} \ \Omega$ must be connected in series with the voltmeter.

Answer

connect resistor of $\frac{x}{2}$ in series with voltmeter.

Answer

connect resistor of $\frac{x}{2}$ in parallel to voltmeter.

Answer

connect resistor of $x \ \Omega$ in series with voltmeter.

Answer

connect resistor of $2x \ \Omega$ in parallel to voltmeter.

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