Moving Coil Galvanometer:

• Force experienced by the coil due to magnetic field is given by the equation: $$F = BIL\sin(\theta)$$
• Torque experienced by the coil is given by the equation: $$\tau = BIL\sin(\theta) \cdot N \cdot A$$
• The restoring torque is proportional to the current passing through the coil

Ammeter:

• The galvanometer is converted into an ammeter by connecting a shunt resistor in parallel
• The shunt resistor allows only a fraction of the total current to pass through the galvanometer

Voltmeter:

• The galvanometer is converted into a voltmeter by connecting a series resistor
• The series resistor limits the current passing through the galvanometer

• The current passing through the shunt resistor: $$I_{\text{shunt}} = \frac{I - IG}{R_{\text{shunt}}/N_{\text{shunt}}A}$$
• The current passing through the galvanometer: $$IG = \frac{IR_{\text{shunt}}}{N_{\text{shunt}}R_{\text{g}}+R_{\text{shunt}}}$$
• Total current passing through the ammeter: $$I = I_{\text{shunt}} + IG$$

• A 0.1 Ω shunt resistor with 10 turns and 1 mm² area is connected to a galvanometer having a resistance of 50 Ω.
• If the current is 2 A, calculate the current passing through the galvanometer.
• Solution: $$IG = \frac{2 \times 0.1}{10 \times 1 \times 10^{-6} \times 50+0.1} = 0.00198 , \text{A}$$

• The voltage across the series resistor: $$V_{\text{series}} = I_g (R_{\text{series}} + R_{\text{g}})$$
• The voltage across the galvanometer: $$I_g R_{\text{g}} = \frac{R_{\text{series}} + R_{\text{g}}}{R_{\text{series}}} V_{\text{series}}$$
• Total voltage across the voltmeter: $$V = V_{\text{series}} + I_g R_{\text{g}}$$

• A voltmeter is constructed using a galvanometer with a resistance of 50 Ω. A series resistor of 1000 Ω is connected.
• If a voltage of 5 V is applied across the voltmeter, calculate the voltage across the galvanometer.
• Solution: $$I_g R_{\text{g}} = \frac{1000 + 50}{1000} \times 5 , \text{V} = 5.25 , \text{V}$$

• When the dipole is aligned with an external electric field, work is done to bring the charges closer together.
• The potential energy is negative, implying that energy is released when the dipole interacts with the field.

• If the dipole is aligned perpendicular to the electric field, the equation becomes: $$U = -\frac{pE}{r^2}$$

• If the dipole is at an angle θ with the field, the equation becomes: $$U = -pE \cos(\theta)$$

• An electric dipole with a dipole moment of 10 C.m is placed in an electric field of magnitude 100 N/C.
• If the angle between the dipole and field is 30 degrees, calculate the potential energy of the dipole.
• Solution: $$U = -10 \times 100 \times \cos(30) = -500 , \text{J}$$

• The torque tends to align the dipole with the electric field.
• The torque is maximum when the dipole is perpendicular to the field.

• An electric dipole with a dipole moment of 5 C.m is placed in an electric field of magnitude 200 N/C.
• If the angle between the dipole and field is 45 degrees, calculate the torque experienced by the dipole.
• Solution: $$\tau = 5 \times 200 \times \sin(45) = 500 , \text{N.m}$$

• The potential energy is positive, indicating the potential energy required to bring the charges together.

• Three charges q1 = 2 µC, q2 = -3 µC, and q3 = 5 µC are located at the vertices of an equilateral triangle with sides of 2 m.
• Find the potential energy of the system of point charges.
• Solution: $$U = \frac{1}{4\pi\epsilon_0}\left(\frac{2\times(-3)}{2} + \frac{2\times5}{2} + \frac{(-3)\times5}{2}\right) , \text{J}$$

• Two capacitors with capacitance values of 5 µF and 10 µF are connected in series.
• Calculate the equivalent capacitance of the system.
• Solution: $$\frac{1}{C_{\text{eq}}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$$

• Two capacitors with capacitance values of 5 µF and 10 µF are connected in parallel.
• Calculate the equivalent capacitance of the system.
• Solution: $$C_{\text{eq}} = 5 + 10 = 15 , \mu\text{F}$$

• Length: Longer the wire, higher the resistance
• Cross-sectional area: Smaller the area, higher the resistance
• Temperature: Higher the temperature, higher the resistance
• Material: Different materials have different resistivities
• Resistivity: Intrinsic property of a material that determines its resistance

• Voltage (V) is directly proportional to current (I) if the resistance (R) is constant.
• Voltage is inversely proportional to resistance if the current is constant.
• Current is directly proportional to voltage if resistance is constant.

• Two resistors with resistance values of 5 Ω and 10 Ω are connected in series.
• Calculate the equivalent resistance of the system.
• Solution: $$R_{\text{eq}} = 5 + 10 = 15 , \Omega$$

• Two resistors with resistance values of 5 Ω and 10 Ω are connected in parallel.
• Calculate the equivalent resistance of the system.
• Solution: $$\frac{1}{R_{\text{eq}}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$$

• The algebraic sum of currents entering any junction in a circuit is zero.

• The sum of potential differences around any closed loop in a circuit is equal to zero.

• In a circuit, three currents enter a junction: 2 A, 3 A, and 5 A. Determine the current leaving the junction.
• Solution: Since current entering and leaving a junction must balance, the current leaving the junction is 10 A.

• Three identical cells with individual voltages of 1.5 V each are connected in series and parallel.
• Calculate the total voltage in each case.
• Solution: In the series connection, the total voltage is 4.5 V (1.5 V + 1.5 V + 1.5 V). In the parallel connection, the total voltage is 1.5 V.

• Electric power is directly proportional to current and voltage.
• Power is the rate of energy transfer or energy consumed per unit time.

• In a circuit, a current of 3 A flows through a resistor with a voltage drop of 5 V. Calculate the power dissipated by the resistor.
• Solution: $$P = 3 \times 5 = 15 , \text{W}$$

• The heat produced is directly proportional to the square of the current, resistance, and time.
• Heat is dissipated as a result of the flow of electric current.

• A current of 2 A flows through a resistor of 5 Ω for 5 minutes. Calculate the heat produced by the resistor.
• Solution: $$H = 2^2 \times 5 \times 5 \times 60 = 12000 , \text{J}$$

• Focal length is the distance between the mirror and its focal point.
• The distance of the object (u) and image (v) from the mirror are used to calculate the focal length.

• A spherical mirror forms an image at a distance of 10 cm when the object is placed at a distance of 15 cm from the mirror. Calculate the focal length of the mirror.
• Solution: $$\frac{1}{f} = \frac{1}{10} + \frac{1}{15}, f = 30 , \text{cm}$$

• The object distance (u) and image distance (v) are used to calculate the focal length (f).
• The focal length can be positive or negative depending on the lens type.

• A lens forms an image at a distance of 30 cm when the object is placed at a distance of 20 cm from the lens. Calculate the focal length of the lens.
• Solution: $$\frac{1}{f} = \frac{1}{30} - \frac{1}{20}, f = -60 , \text{cm}$$

• The force is perpendicular to both the magnetic field and the current direction
• The magnitude of the force increases with the current, magnetic field strength, and length of the conductor
• The force can be determined using the right-hand rule for direction

• The force is perpendicular to both the velocity and the magnetic field
• The magnitude of the force increases with the charge, velocity, and magnetic field strength
• The force can be determined using the right-hand rule for direction

• The magnetic field is directly proportional to the current and inversely proportional to the radius of the loop
• The direction of the magnetic field can be determined using the right-hand rule

• The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions
• The magnitude of the force increases with the currents, length of the conductors, and decreases with the distance between them