Question

The value of the current sensitivity of a galvanometer is given by:

• A. \(\frac{k}{NBA}\)
• B. \(\frac{NBA}{k}\)
• C. \(\frac{kBA}{N}\)
• D. \(\frac{kNB}{A}\)
• A. 0.25 \(\Omega\)
• B. 0.30 \(\Omega\)
• C. 0.50 \(\Omega\)
• D. 6.0 \(\Omega\)
• A. 0.20 \(\Omega\)
• B. 0.24 \(\Omega\)
• C. 6.0 \(\Omega\)
• D. 6.25 \(\Omega\)
• A. \(R_2 - 2R_1\)
• B. \(R_2 - R_1\)
• C. \(R_1 + R_2\)
• D. \(R_1 - 2R_2\)
• A. \(3.6 \times 10^{-3}\) Nm
• B. \(1.8 \times 10^{-4}\) Nm
• C. \(2.4 \times 10^{-3}\) Nm
• D. \(1.2 \times 10^{-4}\) Nm

Answer 1

The Correct Option is B

The current sensitivity of a galvanometer is defined as the deflection produced per unit current, and the expression for it is: \[ \text{Current Sensitivity} = \frac{NBA}{k} \] Where:
- \(k\) is a constant that depends on the galvanometer,
- \(N\) is the number of turns in the coil,
- \(B\) is the magnetic field,
- \(A\) is the area of the coil. Thus, the correct option is \(\frac{NBA}{k}\).

Answer 2

To convert a galvanometer into an ammeter, a shunt resistor is connected in parallel with the galvanometer. The value of the shunt resistor is calculated using the following formula: \[ I_{\text{max}} = \frac{V_{\text{g}}}{R_{\text{g}}} \] Where: - \(I_{\text{max}}\) is the full-scale current for the ammeter, - \(V_{\text{g}}\) is the voltage across the galvanometer at full scale, - \(R_{\text{g}}\) is the resistance of the galvanometer. For a galvanometer with resistance \(R_{\text{g}} = 6 \, \Omega\) and full-scale deflection current \(I_{\text{g}} = 0.2 \, A\), the voltage across the galvanometer is: \[ V_{\text{g}} = I_{\text{g}} \cdot R_{\text{g}} = 0.2 \times 6 = 1.2 \, \text{V} \] Now, to convert this galvanometer into an ammeter of range (0 – 5 A), the voltage across the galvanometer must remain the same, and the current that passes through the shunt resistor should be: \[ I_{\text{max}} = 5 \, A \] The current through the shunt resistor, \(I_{\text{s}}\), will be: \[ I_{\text{s}} = I_{\text{max}} - I_{\text{g}} = 5 - 0.2 = 4.8 \, A \] The value of the shunt resistor \(R_{\text{s}}\) can be calculated using Ohm’s law: \[ R_{\text{s}} = \frac{V_{\text{g}}}{I_{\text{s}}} = \frac{1.2}{4.8} = 0.25 \, \Omega \] Thus, the value of the shunt resistor is 0.25 \(\Omega\).

Answer 3

In part (ii), we calculated the value of the shunt resistor to be 0.25 \(\Omega\). Now, we need to calculate the total resistance of the ammeter, which consists of the galvanometer resistance and the shunt resistor in parallel. The total resistance \(R_{\text{total}}\) of the ammeter is given by the parallel combination of the galvanometer resistance \(R_{\text{g}}\) and the shunt resistance \(R_{\text{s}}\): \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{g}}} + \frac{1}{R_{\text{s}}} \] Substituting the values \(R_{\text{g}} = 6 \, \Omega\) and \(R_{\text{s}} = 0.25 \, \Omega\): \[ \frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{0.25} = \frac{1}{6} + 4 \] \[ \frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{24}{6} = \frac{25}{6} \] \[ R_{\text{total}} = \frac{6}{25} = 0.24 \, \Omega \] Thus, the total resistance of the ammeter is \(0.24 \, \Omega\).

Answer 4

The resistance of a galvanometer is \(R_g\), and to convert it into a voltmeter, we add a series resistance. The range of the voltmeter is given by: \[ V = I_g \cdot (R_g + R) \] Where: 
- \(V\) is the range of the voltmeter, 
- \(I_g\) is the current at full scale deflection for the galvanometer, 
- \(R_g\) is the resistance of the galvanometer, 
- \(R\) is the series resistance added. For range \(0 - V\), the series resistance is \(R_1\), so the total resistance is \(R_g + R_1\). For range \(0 - 2V\), the series resistance is \(R_2\), so the total resistance is \(R_g + R_2\). Since the voltage is doubled when \(R_1\) is replaced by \(R_2\), we have: \[ \frac{R_g + R_2}{R_g + R_1} = 2 \] Solving for \(R_g\): \[ R_g + R_2 = 2(R_g + R_1) \] \[ R_g + R_2 = 2R_g + 2R_1 \] \[ R_2 = R_g + 2R_1 \] Thus, the resistance of the galvanometer is: \[ R_g = R_2 - 2R_1 \] Therefore, the correct answer is \(R_2 - 2R_1\).

Answer 5

The deflecting torque \(T\) on a coil in a magnetic field is given by the formula: \[ T = n B A I \] Where: 
- \(n\) is the number of turns, 
- \(B\) is the magnetic field strength, 
- \(A\) is the area of the coil, 
- \(I\) is the current flowing through the coil. Substituting the given values: 
- \(n = 100\), 
- \(B = 0.20 \, \text{T}\), 
- \(A = 18 \, \text{cm}^2 = 18 \times 10^{-4} \, \text{m}^2\), - \(I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\), The torque is: \[ T = 100 \times 0.20 \times 18 \times 10^{-4} \times 5 \times 10^{-3} \] \[ T = 1.8 \times 10^{-4} \, \text{Nm} \] Thus, the deflecting torque acting on the coil is \(1.8 \times 10^{-4} \, \text{Nm}\).