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 per unit current. Its expression is: \[ \text{Current Sensitivity} = \frac{NBA}{k} \] Where:
- \(k\) represents a galvanometer-dependent constant.
- \(N\) is the number of coil turns.
- \(B\) denotes the magnetic field.
- \(A\) signifies the coil's area. Therefore, the accurate expression for current sensitivity is \(\frac{NBA}{k}\).

Answer 2

A galvanometer can be transformed into an ammeter by connecting a shunt resistor in parallel with it. The required shunt resistance is determined by the formula: \[ I_{\text{max}} = \frac{V_{\text{g}}}{R_{\text{g}}} \] where \(I_{\text{max}}\) represents the ammeter's full-scale current, \(V_{\text{g}}\) is the voltage across the galvanometer at full scale, and \(R_{\text{g}}\) is the galvanometer's resistance. For a galvanometer with \(R_{\text{g}} = 6 \, \Omega\) and a full-scale deflection current of \(I_{\text{g}} = 0.2 \, A\), the voltage across the galvanometer is calculated as: \[ V_{\text{g}} = I_{\text{g}} \cdot R_{\text{g}} = 0.2 \times 6 = 1.2 \, \text{V} \] To modify this galvanometer into an ammeter with a range of 0 – 5 A, the voltage across the galvanometer must remain constant. The current through the shunt resistor should then be: \[ I_{\text{max}} = 5 \, A \] The current flowing through the shunt resistor, \(I_{\text{s}}\), is given by: \[ I_{\text{s}} = I_{\text{max}} - I_{\text{g}} = 5 - 0.2 = 4.8 \, A \] Ohm's law is used to compute the value of the shunt resistor, \(R_{\text{s}}\): \[ R_{\text{s}} = \frac{V_{\text{g}}}{I_{\text{s}}} = \frac{1.2}{4.8} = 0.25 \, \Omega \] Consequently, the necessary shunt resistor value is 0.25 \(\Omega\).

Answer 3

Following the calculation in part (ii) which determined the shunt resistor value to be \(0.25 \, \Omega\), the subsequent step is to compute the ammeter's total resistance. This total resistance is a result of the galvanometer resistance and the shunt resistor connected in parallel. The formula for the total resistance \(R_{\text{total}}\) of the ammeter, representing the parallel combination of galvanometer resistance \(R_{\text{g}}\) and shunt resistance \(R_{\text{s}}\), is: \[\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{g}}} + \frac{1}{R_{\text{s}}}\] Substituting the given values, \(R_{\text{g}} = 6 \, \Omega\) and \(R_{\text{s}} = 0.25 \, \Omega\): \[\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{0.25} = \frac{1}{6} + 4\] \[\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{24}{6} = \frac{25}{6}\] Consequently, the total resistance is: \[R_{\text{total}} = \frac{6}{25} = 0.24 \, \Omega\] Therefore, the ammeter's total resistance is \(0.24 \, \Omega\).

Answer 4

A galvanometer with resistance \(R_g\) can be converted into a voltmeter by adding a series resistance. The voltmeter's range \(V\) is determined by the formula: \[ V = I_g \cdot (R_g + R) \] where \(V\) is the voltmeter's range, \(I_g\) is the galvanometer's full-scale deflection current, \(R_g\) is the galvanometer's resistance, and \(R\) is the added series resistance. For a range of \(0\) to \(V\), the series resistance is \(R_1\), resulting in a total resistance of \(R_g + R_1\). For a range of \(0\) to \(2V\), the series resistance is \(R_2\), resulting in a total resistance of \(R_g + R_2\). Since doubling the range (from \(V\) to \(2V\)) occurs when \(R_1\) is replaced by \(R_2\), the following relationship holds: \[ \frac{R_g + R_2}{R_g + R_1} = 2 \] This equation can be solved for \(R_g\): \[ R_g + R_2 = 2(R_g + R_1) \] \[ R_g + R_2 = 2R_g + 2R_1 \] \[ R_2 = R_g + 2R_1 \] Therefore, the resistance of the galvanometer is calculated as: \[ R_g = R_2 - 2R_1 \] The correct answer is \(R_2 - 2R_1\).

Answer 5

The deflecting torque \(T\) on a coil in a magnetic field is calculated using the formula: \[ T = n B A I \]. Here, \(n\) represents the number of turns, \(B\) is the magnetic field strength, \(A\) is the coil's area, and \(I\) is the current through the coil. Given the values: \(n = 100\), \(B = 0.20 \, \text{T}\), \(A = 18 \, \text{cm}^2 = 18 \times 10^{-4} \, \text{m}^2\), and \(I = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\), the torque is computed as: \[ T = 100 \times 0.20 \times 18 \times 10^{-4} \times 5 \times 10^{-3} \] This results in a torque of \[ T = 1.8 \times 10^{-4} \, \text{Nm} \]. Consequently, the deflecting torque acting on the coil is \(1.8 \times 10^{-4} \, \text{Nm}\).