Q184 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
The range of voltmeter is $10\ \text{V}$ and its resistance is $50\ \Omega$. To convert it to a voltmeter of range $15\ \text{V}$, how much resistance is to be added?
- A$25\ \Omega$ resistor in parallel.
- Correct answerB$25\ \Omega$ resistor in series.
- C$125\ \Omega$ resistor in parallel.
- D$125\ \Omega$ resistor in series.
Solution
Current for full scale deflection of the voltmeter is given by $I_g = \dfrac{V_1}{R_v}$.
Substituting the given values, $I_g = \dfrac{10}{50} = 0.2\ \text{A}$.
To increase the range of the voltmeter to $V_2 = 15\ \text{V}$, a resistance $R$ must be connected in series with it.
The new range is given by $V_2 = I_g(R_v + R)$.
Substituting the values, $15 = 0.2(50 + R)$.
$50 + R = \dfrac{15}{0.2} = 75$
$R = 75 - 50 = 25\ \Omega$.
Thus, a $25\ \Omega$ resistor must be connected in series.
Answer: $25\ \Omega$ resistor in series.
Substituting the given values, $I_g = \dfrac{10}{50} = 0.2\ \text{A}$.
To increase the range of the voltmeter to $V_2 = 15\ \text{V}$, a resistance $R$ must be connected in series with it.
The new range is given by $V_2 = I_g(R_v + R)$.
Substituting the values, $15 = 0.2(50 + R)$.
$50 + R = \dfrac{15}{0.2} = 75$
$R = 75 - 50 = 25\ \Omega$.
Thus, a $25\ \Omega$ resistor must be connected in series.
Answer: $25\ \Omega$ resistor in series.
Hence, the correct answer is option: 'B'.