Questions & explanations
1. A battery has an emf of 10 V and an internal resistance of 5 Ω. It is connected to an external variable resistor R. What is the maximum power that the battery can deliver to R?
- 5 W
- 10 W
- 20 W
- 2.5 W
Answer: 5 W
For maximum power transfer from a source with internal resistance r to an external load R, the condition is R = r. Here r = 5 Ω, so the load resistance should be 5 Ω. The total resistance is then 10 Ω, and the current is I = E / (r + R) = 10 V / 10 Ω = 1 A. The power delivered to the load is P = I²R = (1 A)² × 5 Ω = 5 W. Alternatively, using the formula P_max = E² / (4r) = (10 V)² / (4 × 5 Ω) = 100 / 20 = 5 W. The other options are incorrect because they do not correspond to the power obtained under the maximum power transfer condition.
2. The electromotive force (emf) of a cell is defined as:
- The potential difference between its terminals when it is delivering current
- The potential difference between its terminals when no current is drawn from it
- The total energy supplied by the cell per unit charge when it is connected to an external circuit
- The product of the current and the external resistance
Answer: The potential difference between its terminals when no current is drawn from it
The emf of a cell is the potential difference across its terminals in an open circuit (when no current flows). It represents the energy supplied per unit charge by the cell. Option (A) is the terminal voltage under load, which is less than emf due to internal resistance. Option (C) is close but not exactly the definition; emf is the open-circuit terminal voltage, not the energy when connected. Option (D) is the voltage across the external resistor.
3. Why is an ammeter always connected in series in a circuit and a voltmeter always connected in parallel?
- An ammeter has very low resistance and must carry the entire current; a voltmeter has very high resistance and must measure potential difference across two points without drawing significant current.
- An ammeter has very high resistance and must be inserted in the path; a voltmeter has very low resistance and must be placed across the component.
- An ammeter measures voltage and is therefore connected across; a voltmeter measures current and is therefore connected in series.
- An ammeter and a voltmeter are interchangeable; the connection depends only on the magnitude of current or voltage being measured.
Answer: An ammeter has very low resistance and must carry the entire current; a voltmeter has very high resistance and must measure potential difference across two points without drawing significant current.
An ammeter is used to measure current, so it must be placed in series so that the entire current flows through it. Its low resistance ensures minimal voltage drop. A voltmeter measures potential difference, so it is connected in parallel across the component. Its high resistance ensures it draws negligible current, thus not disturbing the circuit. The other options reverse the roles or suggest interchangeability, which is incorrect.
4. Consider a circuit with two loops: A 12 V battery (positive terminal upward) is connected in series with a 2 Ω resistor and a 4 Ω resistor. A 6 V battery (positive terminal upward) is connected in series with a 2 Ω resistor and the same 4 Ω resistor. The 4 Ω resistor is common to both branches. Using Kirchhoff's laws, the magnitude of the current flowing through the 4 Ω resistor is:
- 0.2 A
- 0.4 A
- 0.6 A
- 0.8 A
Answer: 0.6 A
Let I₁ (clockwise) be the current in the left loop and I₂ (clockwise) in the right loop. Applying Kirchhoff's loop rule: Left loop: 12 = 2I₁ + 4(I₁ - I₂) → 12 = 6I₁ - 4I₂. Right loop: 6 = 2I₂ + 4(I₂ - I₁) → 6 = 6I₂ - 4I₁. Solving: multiply first by 3: 36 = 18I₁ - 12I₂; multiply second by 2: 12 = 12I₂ - 8I₁. Adding gives 48 = 10I₁ → I₁ = 4.8 A. Substituting back gives I₂ = 4.2 A. Current through the 4 Ω resistor = I₁ - I₂ = 0.6 A.
5. A moving coil galvanometer works on the principle that:
- A current-carrying coil placed in a uniform radial magnetic field experiences a torque directly proportional to the current, and the deflection is indicated by a pointer attached to the coil.
- A stationary coil placed in a changing magnetic field induces a current that deflects a pointer.
- The force between two parallel current-carrying conductors produces a deflection proportional to the current.
- The magnetic field due to a current-carrying wire exerts a force on a permanent magnet, causing deflection.
Answer: A current-carrying coil placed in a uniform radial magnetic field experiences a torque directly proportional to the current, and the deflection is indicated by a pointer attached to the coil.
The moving coil galvanometer uses the torque on a current-carrying coil in a radial magnetic field. The torque is τ = N I A B, which is proportional to current I. A spring provides a restoring torque, so the deflection θ is proportional to I. The other options describe different phenomena (electromagnetic induction, force between conductors, or magnetic force on a magnet) that are not the operating principle of this device.
6. To convert a moving coil galvanometer into an ammeter that can measure a larger current, one should connect:
- A low resistance in parallel with the galvanometer coil.
- A high resistance in series with the galvanometer coil.
- A low resistance in series with the galvanometer coil.
- A high resistance in parallel with the galvanometer coil.
Answer: A low resistance in parallel with the galvanometer coil.
An ammeter is a low-resistance device. To convert a galvanometer into an ammeter, a low-resistance shunt is connected in parallel so that most of the current bypasses the galvanometer, allowing the meter to measure currents much larger than its full-scale deflection current. The other connections (high series, low series, or high parallel) would not achieve the desired range extension or would damage the galvanometer.
7. An ideal ammeter should have ______ resistance, and an ideal voltmeter should have ______ resistance.
- zero; infinite
- infinite; zero
- low; low
- high; high
Answer: zero; infinite
An ideal ammeter is connected in series and must not impede the current, so its resistance should be zero. An ideal voltmeter is connected in parallel and should not draw any current from the circuit, so its resistance should be infinite. The other choices are opposite or incorrect because low resistance in a voltmeter would cause loading, and high resistance in an ammeter would reduce the measured current.
8. In a balanced Wheatstone bridge, the resistances in the four arms (in order) are P, Q, R, and S. The condition for balance is:
- P/Q = R/S
- P/R = Q/S
- P/Q = S/R
- P + S = Q + R
Answer: P/Q = R/S
The Wheatstone bridge is balanced when no current flows through the galvanometer. Under this condition, the potential difference across the two arms are equal, leading to the ratio P/Q = R/S. Option (B) is incorrect because it implies P/R = Q/S, which is a different arrangement. Option (C) reverses the ratio. Option (D) is a condition for a different type of network.
9. Two identical cells, each of emf E and internal resistance r, are connected in parallel. The equivalent emf and internal resistance of the combination are:
- E, r/2
- 2E, r/2
- E, 2r
- 2E, 2r
Answer: E, r/2
For identical cells in parallel, the equivalent emf is the same as that of one cell (E) because they share the same potential difference. The equivalent internal resistance becomes r/n = r/2, as the effective cross-sectional area for current flow increases. Options (B) and (D) would be for series combination, and (C) has wrong internal resistance.
10. To convert a moving coil galvanometer into a voltmeter that can measure a larger voltage, one should connect:
- A high resistance in series with the galvanometer coil.
- A low resistance in parallel with the galvanometer coil.
- A high resistance in parallel with the galvanometer coil.
- A low resistance in series with the galvanometer coil.
Answer: A high resistance in series with the galvanometer coil.
A voltmeter must have high resistance. By connecting a large series resistance, the galvanometer can measure a larger potential difference without drawing excessive current. The other connections (low parallel, high parallel, low series) would either short the circuit, not limit current, or fail to provide the required high input resistance.
11. Which of the following statements about Ohm's law is correct?
- All conductors obey Ohm's law at all temperatures
- For ohmic conductors, the ratio V/I is constant at a given temperature
- Ohm's law states that V ∝ I for all materials
- Non-ohmic conductors also obey Ohm's law but only at low voltages
Answer: For ohmic conductors, the ratio V/I is constant at a given temperature
Ohm's law is obeyed by ohmic conductors (e.g., metals) only when temperature and other physical conditions are constant; the V-I graph is linear. Non-ohmic conductors (e.g., semiconductors, diodes) do not follow Ohm's law. The other options falsely claim all conductors obey it or that non-ohmic conductors obey it under certain conditions.
12. A galvanometer has a resistance of 50 Ω and gives full-scale deflection with a current of 10 mA. What shunt resistance is needed to convert it into an ammeter of range 1 A?
- 0.505 Ω
- 5.05 Ω
- 50.5 Ω
- 0.05 Ω
Answer: 0.505 Ω
The shunt resistance S = (I_g × G) / (I - I_g), where I_g = 0.01 A, G = 50 Ω, I = 1 A. So S = (0.01 × 50) / (1 - 0.01) = 0.5 / 0.99 ≈ 0.505 Ω. This low resistance in parallel diverts most of the current away from the galvanometer, allowing the meter to measure up to 1 A. The other options are orders of magnitude off or miscalculated.