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Current Electricity
Physics XII - Chapter 03: Complete NCERT Exercise Solutions
Master Current Electricity with NCERT solutions for Ohm's law, Kirchhoff's laws, electrical networks, and measuring instruments.
Question 3.1
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Answer & Explanation:
Step 1: Maximum current occurs when external resistance is zero (short circuit).
Using Ohm's law: \( I_{\max} = \frac{\text{emf}}{r} \)
\( I_{\max} = \frac{12}{0.4} = 30 \, \text{A} \).
Answer: 30 A.
Using Ohm's law: \( I_{\max} = \frac{\text{emf}}{r} \)
\( I_{\max} = \frac{12}{0.4} = 30 \, \text{A} \).
Answer: 30 A.
Question 3.2
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer & Explanation:
Step 1: Find total resistance
\( I = \frac{E}{R + r} \) → \( 0.5 = \frac{10}{R + 3} \)
\( R + 3 = 20 \) → \( R = 17 \, \Omega \).
Step 2: Terminal voltage
\( V = E - I r = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5 \, \text{V} \).
Answer: Resistance = 17 Ω, Terminal voltage = 8.5 V.
\( I = \frac{E}{R + r} \) → \( 0.5 = \frac{10}{R + 3} \)
\( R + 3 = 20 \) → \( R = 17 \, \Omega \).
Step 2: Terminal voltage
\( V = E - I r = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5 \, \text{V} \).
Answer: Resistance = 17 Ω, Terminal voltage = 8.5 V.
Question 3.3
(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer & Explanation:
(a) Series total: \( R_s = 1 + 2 + 3 = 6 \, \Omega \).
(b) Current: \( I = \frac{V}{R_s} = \frac{12}{6} = 2 \, \text{A} \).
Potential drops: \( V_1 = I R_1 = 2 \times 1 = 2 \, \text{V} \)
\( V_2 = 2 \times 2 = 4 \, \text{V} \)
\( V_3 = 2 \times 3 = 6 \, \text{V} \).
Answer: (a) 6 Ω, (b) 2 V, 4 V, 6 V.
(b) Current: \( I = \frac{V}{R_s} = \frac{12}{6} = 2 \, \text{A} \).
Potential drops: \( V_1 = I R_1 = 2 \times 1 = 2 \, \text{V} \)
\( V_2 = 2 \times 2 = 4 \, \text{V} \)
\( V_3 = 2 \times 3 = 6 \, \text{V} \).
Answer: (a) 6 Ω, (b) 2 V, 4 V, 6 V.
Question 3.4
(a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Answer & Explanation:
(a) Parallel total: \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} = 0.5 + 0.25 + 0.2 = 0.95 \)
\( R_p = \frac{1}{0.95} \approx 1.053 \, \Omega \).
(b) Voltage across each = 20 V.
Currents: \( I_1 = \frac{20}{2} = 10 \, \text{A} \), \( I_2 = \frac{20}{4} = 5 \, \text{A} \), \( I_3 = \frac{20}{5} = 4 \, \text{A} \).
Total current: \( I_{\text{total}} = 10 + 5 + 4 = 19 \, \text{A} \).
Answer: (a) ≈1.053 Ω, (b) 10 A, 5 A, 4 A; total 19 A.
\( R_p = \frac{1}{0.95} \approx 1.053 \, \Omega \).
(b) Voltage across each = 20 V.
Currents: \( I_1 = \frac{20}{2} = 10 \, \text{A} \), \( I_2 = \frac{20}{4} = 5 \, \text{A} \), \( I_3 = \frac{20}{5} = 4 \, \text{A} \).
Total current: \( I_{\text{total}} = 10 + 5 + 4 = 19 \, \text{A} \).
Answer: (a) ≈1.053 Ω, (b) 10 A, 5 A, 4 A; total 19 A.
Question 3.5
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is \(1.70 \times 10^{-4} \, \text{°C}^{-1}\).
Answer & Explanation:
Formula: \( R_t = R_0 [1 + \alpha (t - t_0)] \)
\( 117 = 100 [1 + 1.70 \times 10^{-4} (t - 27)] \)
\( 1.17 = 1 + 1.70 \times 10^{-4} (t - 27) \)
\( 0.17 = 1.70 \times 10^{-4} (t - 27) \)
\( t - 27 = \frac{0.17}{1.70 \times 10^{-4}} = 1000 \)
\( t = 1027 \, \text{°C} \).
Answer: 1027 °C.
\( 117 = 100 [1 + 1.70 \times 10^{-4} (t - 27)] \)
\( 1.17 = 1 + 1.70 \times 10^{-4} (t - 27) \)
\( 0.17 = 1.70 \times 10^{-4} (t - 27) \)
\( t - 27 = \frac{0.17}{1.70 \times 10^{-4}} = 1000 \)
\( t = 1027 \, \text{°C} \).
Answer: 1027 °C.
Question 3.6
A negligibly small current is passed through a wire of length 15 m and uniform cross-section \(6.0 \times 10^{-7} \, \text{m}^2\), and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer & Explanation:
Using \( R = \rho \frac{l}{A} \):
\( \rho = \frac{R A}{l} = \frac{5.0 \times 6.0 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \, \Omega\cdot\text{m} \).
Answer: \( 2.0 \times 10^{-7} \, \Omega\cdot\text{m} \).
\( \rho = \frac{R A}{l} = \frac{5.0 \times 6.0 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \, \Omega\cdot\text{m} \).
Answer: \( 2.0 \times 10^{-7} \, \Omega\cdot\text{m} \).
Question 3.7
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Answer & Explanation:
Using \( R_t = R_0 [1 + \alpha (t - t_0)] \):
\( 2.7 = 2.1 [1 + \alpha (100 - 27.5)] \)
\( \frac{2.7}{2.1} = 1 + \alpha \times 72.5 \)
\( 1.2857 = 1 + 72.5 \alpha \)
\( 0.2857 = 72.5 \alpha \)
\( \alpha = \frac{0.2857}{72.5} \approx 3.94 \times 10^{-3} \, \text{°C}^{-1} \).
Answer: \( 3.94 \times 10^{-3} \, \text{°C}^{-1} \).
\( 2.7 = 2.1 [1 + \alpha (100 - 27.5)] \)
\( \frac{2.7}{2.1} = 1 + \alpha \times 72.5 \)
\( 1.2857 = 1 + 72.5 \alpha \)
\( 0.2857 = 72.5 \alpha \)
\( \alpha = \frac{0.2857}{72.5} \approx 3.94 \times 10^{-3} \, \text{°C}^{-1} \).
Answer: \( 3.94 \times 10^{-3} \, \text{°C}^{-1} \).
Question 3.8
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is \(1.70 \times 10^{-4} \, \text{°C}^{-1}\).
Answer & Explanation:
Step 1: Resistance at room temp
\( R_0 = \frac{V}{I_0} = \frac{230}{3.2} = 71.875 \, \Omega \).
Step 2: Resistance at steady temp
\( R_t = \frac{V}{I_t} = \frac{230}{2.8} = 82.143 \, \Omega \).
Step 3: Find temperature
\( R_t = R_0 [1 + \alpha (t - t_0)] \)
\( 82.143 = 71.875 [1 + 1.70 \times 10^{-4} (t - 27)] \)
\( 1.143 = 1 + 1.70 \times 10^{-4} (t - 27) \)
\( 0.143 = 1.70 \times 10^{-4} (t - 27) \)
\( t - 27 = \frac{0.143}{1.70 \times 10^{-4}} \approx 841.18 \)
\( t \approx 868.18 \, \text{°C} \).
Answer: ≈ 868 °C.
\( R_0 = \frac{V}{I_0} = \frac{230}{3.2} = 71.875 \, \Omega \).
Step 2: Resistance at steady temp
\( R_t = \frac{V}{I_t} = \frac{230}{2.8} = 82.143 \, \Omega \).
Step 3: Find temperature
\( R_t = R_0 [1 + \alpha (t - t_0)] \)
\( 82.143 = 71.875 [1 + 1.70 \times 10^{-4} (t - 27)] \)
\( 1.143 = 1 + 1.70 \times 10^{-4} (t - 27) \)
\( 0.143 = 1.70 \times 10^{-4} (t - 27) \)
\( t - 27 = \frac{0.143}{1.70 \times 10^{-4}} \approx 841.18 \)
\( t \approx 868.18 \, \text{°C} \).
Answer: ≈ 868 °C.
Question 3.9
Determine the current in each branch of the network shown in Fig. 3.30.
Answer & Explanation:
Network analysis:
Apply Kirchhoff's laws.
Let currents in branches be \(I_1, I_2, I_3\) etc.
Solve simultaneous equations.
Detailed solution requires figure values.
Apply Kirchhoff's laws.
Let currents in branches be \(I_1, I_2, I_3\) etc.
Solve simultaneous equations.
Detailed solution requires figure values.
Question 3.10
(a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Answer & Explanation:
(a) Using balance condition: \( \frac{X}{Y} = \frac{l}{100-l} \)
\( \frac{X}{12.5} = \frac{39.5}{60.5} \) → \( X = 12.5 \times \frac{39.5}{60.5} \approx 8.16 \, \Omega \).
Thick copper strips minimize resistance of connecting wires.
(b) If X and Y interchanged: \( \frac{Y}{X} = \frac{l'}{100-l'} \) → solve for new l'.
(c) Interchanging galvanometer and cell: Balance point unchanged, galvanometer shows zero current.
Answer: (a) ≈8.16 Ω, (b) l' = 60.5 cm, (c) no current.
\( \frac{X}{12.5} = \frac{39.5}{60.5} \) → \( X = 12.5 \times \frac{39.5}{60.5} \approx 8.16 \, \Omega \).
Thick copper strips minimize resistance of connecting wires.
(b) If X and Y interchanged: \( \frac{Y}{X} = \frac{l'}{100-l'} \) → solve for new l'.
(c) Interchanging galvanometer and cell: Balance point unchanged, galvanometer shows zero current.
Answer: (a) ≈8.16 Ω, (b) l' = 60.5 cm, (c) no current.
Question 3.11
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Answer & Explanation:
Step 1: Charging current
\( I = \frac{\text{Supply voltage} - \text{Battery emf}}{R + r} = \frac{120 - 8}{15.5 + 0.5} = \frac{112}{16} = 7 \, \text{A} \).
Step 2: Terminal voltage during charging
\( V = E + I r = 8 + 7 \times 0.5 = 8 + 3.5 = 11.5 \, \text{V} \).
Step 3: Purpose of series resistor
To limit charging current to safe value.
Answer: Terminal voltage = 11.5 V, resistor limits current.
\( I = \frac{\text{Supply voltage} - \text{Battery emf}}{R + r} = \frac{120 - 8}{15.5 + 0.5} = \frac{112}{16} = 7 \, \text{A} \).
Step 2: Terminal voltage during charging
\( V = E + I r = 8 + 7 \times 0.5 = 8 + 3.5 = 11.5 \, \text{V} \).
Step 3: Purpose of series resistor
To limit charging current to safe value.
Answer: Terminal voltage = 11.5 V, resistor limits current.
Question 3.12
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Answer & Explanation:
Using proportionality: \( \frac{E_1}{E_2} = \frac{l_1}{l_2} \)
\( \frac{1.25}{E_2} = \frac{35.0}{63.0} \)
\( E_2 = 1.25 \times \frac{63.0}{35.0} = 2.25 \, \text{V} \).
Answer: 2.25 V.
\( \frac{1.25}{E_2} = \frac{35.0}{63.0} \)
\( E_2 = 1.25 \times \frac{63.0}{35.0} = 2.25 \, \text{V} \).
Answer: 2.25 V.
Question 3.13
The number density of free electrons in a copper conductor estimated in Example 3.1 is \(8.5 \times 10^{28} \, \text{m}^{-3}\). How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is \(2.0 \times 10^{-6} \, \text{m}^{2}\) and it is carrying a current of 3.0 A.
Answer & Explanation:
Step 1: Drift speed \( v_d = \frac{I}{n e A} \)
\( v_d = \frac{3.0}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6}} \)
≈ \( 1.1 \times 10^{-4} \, \text{m/s} \).
Step 2: Time \( t = \frac{L}{v_d} = \frac{3.0}{1.1 \times 10^{-4}} \approx 2.73 \times 10^{4} \, \text{s} \).
Answer: ≈ 2.73 × 10⁴ s (about 7.6 hours).
\( v_d = \frac{3.0}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6}} \)
≈ \( 1.1 \times 10^{-4} \, \text{m/s} \).
Step 2: Time \( t = \frac{L}{v_d} = \frac{3.0}{1.1 \times 10^{-4}} \approx 2.73 \times 10^{4} \, \text{s} \).
Answer: ≈ 2.73 × 10⁴ s (about 7.6 hours).
Question 3.14
The earth's surface has a negative surface charge density of \(10^{-9} \, \text{C} \, \text{m}^{-2}\). The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (Radius of earth = \(6.37 \times 10^6 \, \text{m}\).)
Answer & Explanation:
Step 1: Total surface charge
Surface area of earth \( A = 4\pi R^2 \approx 5.1 \times 10^{14} \, \text{m}^2 \).
Total charge \( Q = \sigma A = 10^{-9} \times 5.1 \times 10^{14} = 5.1 \times 10^{5} \, \text{C} \).
Step 2: Time to neutralize
\( I = 1800 \, \text{A} \), \( t = \frac{Q}{I} = \frac{5.1 \times 10^{5}}{1800} \approx 283.3 \, \text{s} \approx 4.7 \, \text{minutes} \).
Answer: About 4.7 minutes.
Surface area of earth \( A = 4\pi R^2 \approx 5.1 \times 10^{14} \, \text{m}^2 \).
Total charge \( Q = \sigma A = 10^{-9} \times 5.1 \times 10^{14} = 5.1 \times 10^{5} \, \text{C} \).
Step 2: Time to neutralize
\( I = 1800 \, \text{A} \), \( t = \frac{Q}{I} = \frac{5.1 \times 10^{5}}{1800} \approx 283.3 \, \text{s} \approx 4.7 \, \text{minutes} \).
Answer: About 4.7 minutes.
Question 3.15
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Answer & Explanation:
(a) Total emf = 6 × 2 = 12 V, total internal r = 6 × 0.015 = 0.09 Ω.
Current \( I = \frac{12}{8.5 + 0.09} \approx 1.396 \, \text{A} \).
Terminal voltage \( V = I R = 1.396 \times 8.5 \approx 11.87 \, \text{V} \).
(b) Maximum current (short circuit) \( I_{\max} = \frac{1.9}{380} \approx 0.005 \, \text{A} = 5 \, \text{mA} \).
Too small to drive car starter.
Answer: (a) ≈1.4 A, ≈11.87 V; (b) 5 mA, cannot drive starter.
Current \( I = \frac{12}{8.5 + 0.09} \approx 1.396 \, \text{A} \).
Terminal voltage \( V = I R = 1.396 \times 8.5 \approx 11.87 \, \text{V} \).
(b) Maximum current (short circuit) \( I_{\max} = \frac{1.9}{380} \approx 0.005 \, \text{A} = 5 \, \text{mA} \).
Too small to drive car starter.
Answer: (a) ≈1.4 A, ≈11.87 V; (b) 5 mA, cannot drive starter.
Question 3.16
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.
Given: \( \rho_{\text{Al}} = 2.63 \times 10^{-8} \, \Omega\cdot\text{m}, \rho_{\text{Cu}} = 1.72 \times 10^{-8} \, \Omega\cdot\text{m} \), Relative density of Al = 2.7, of Cu = 8.9.
Given: \( \rho_{\text{Al}} = 2.63 \times 10^{-8} \, \Omega\cdot\text{m}, \rho_{\text{Cu}} = 1.72 \times 10^{-8} \, \Omega\cdot\text{m} \), Relative density of Al = 2.7, of Cu = 8.9.
Answer & Explanation:
For same R and length: \( R = \rho \frac{l}{A} \) → \( A \propto \rho \).
Mass \( m = \text{density} \times \text{volume} = d \times A l \).
\( \frac{m_{\text{Al}}}{m_{\text{Cu}}} = \frac{d_{\text{Al}} \rho_{\text{Al}}}{d_{\text{Cu}} \rho_{\text{Cu}}} = \frac{2.7 \times 2.63 \times 10^{-8}}{8.9 \times 1.72 \times 10^{-8}} \approx 0.48 \).
Al wire is lighter.
Hence preferred for overhead cables to reduce weight.
Answer: Aluminium wire is lighter.
Mass \( m = \text{density} \times \text{volume} = d \times A l \).
\( \frac{m_{\text{Al}}}{m_{\text{Cu}}} = \frac{d_{\text{Al}} \rho_{\text{Al}}}{d_{\text{Cu}} \rho_{\text{Cu}}} = \frac{2.7 \times 2.63 \times 10^{-8}}{8.9 \times 1.72 \times 10^{-8}} \approx 0.48 \).
Al wire is lighter.
Hence preferred for overhead cables to reduce weight.
Answer: Aluminium wire is lighter.
Question 3.17
What conclusion can you draw from the following observations on a resistor made of alloy manganin?
Answer & Explanation:
Observations show V/I constant → Ohm's law obeyed.
Manganin has very low temperature coefficient of resistance.
Answer: Manganin obeys Ohm's law and its resistance is nearly independent of temperature.
Manganin has very low temperature coefficient of resistance.
Answer: Manganin obeys Ohm's law and its resistance is nearly independent of temperature.
Question 3.18
Answer the following questions:
(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Answer & Explanation:
(a) Only current is constant.
(b) No, e.g., diodes, transistors, electrolytes.
(c) To deliver high current without significant voltage drop.
(d) To limit current in case of short circuit for safety.
Answer: As above.
(b) No, e.g., diodes, transistors, electrolytes.
(c) To deliver high current without significant voltage drop.
(d) To limit current in case of short circuit for safety.
Answer: As above.
Question 3.19
Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of / increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of \((10^{22}/10^{3})\).
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of / increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of \((10^{22}/10^{3})\).
Answer & Explanation:
(a) greater
(b) lower
(c) nearly independent of
(d) \(10^{22}\)
Answer: (a) greater, (b) lower, (c) nearly independent of, (d) \(10^{22}\).
(b) lower
(c) nearly independent of
(d) \(10^{22}\)
Answer: (a) greater, (b) lower, (c) nearly independent of, (d) \(10^{22}\).
Question 3.20
(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
Answer & Explanation:
(a) (i) All in series: \( R_{\max} = nR \)
(ii) All in parallel: \( R_{\min} = R/n \)
Ratio \( \frac{R_{\max}}{R_{\min}} = n^2 \).
(b) Combine using series/parallel combinations.
(c) Analyze network in figure.
Answer: (a) (i) series, (ii) parallel, ratio \(n^2\).
(ii) All in parallel: \( R_{\min} = R/n \)
Ratio \( \frac{R_{\max}}{R_{\min}} = n^2 \).
(b) Combine using series/parallel combinations.
(c) Analyze network in figure.
Answer: (a) (i) series, (ii) parallel, ratio \(n^2\).
Question 3.21
Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
Answer & Explanation:
Let equivalent resistance of infinite network be \(R_{\text{eq}}\).
Network is self‑similar: \( R_{\text{eq}} = 1 + \frac{1 \cdot R_{\text{eq}}}{1 + R_{\text{eq}}} \).
Solve quadratic: \( R_{\text{eq}}^2 - R_{\text{eq}} - 1 = 0 \) → \( R_{\text{eq}} = \frac{1+\sqrt{5}}{2} \approx 1.618 \, \Omega \).
Total resistance including internal: \( R_{\text{total}} = 1.618 + 0.5 = 2.118 \, \Omega \).
Current \( I = \frac{12}{2.118} \approx 5.67 \, \text{A} \).
Answer: ≈ 5.67 A.
Network is self‑similar: \( R_{\text{eq}} = 1 + \frac{1 \cdot R_{\text{eq}}}{1 + R_{\text{eq}}} \).
Solve quadratic: \( R_{\text{eq}}^2 - R_{\text{eq}} - 1 = 0 \) → \( R_{\text{eq}} = \frac{1+\sqrt{5}}{2} \approx 1.618 \, \Omega \).
Total resistance including internal: \( R_{\text{total}} = 1.618 + 0.5 = 2.118 \, \Omega \).
Current \( I = \frac{12}{2.118} \approx 5.67 \, \text{A} \).
Answer: ≈ 5.67 A.
Question 3.22
Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV? If not, how will you modify the circuit?
(a) What is the value ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV? If not, how will you modify the circuit?
Answer & Explanation:
(a) \( \frac{\varepsilon}{1.02} = \frac{82.3}{67.3} \) → \( \varepsilon \approx 1.247 \, \text{V} \).
(b) Limits current from standard cell, protects it.
(c) No, because it is shorted near balance.
(d) No, potentiometer is independent of driver cell internal resistance if current constant.
(e) Yes, if potential gradient still sufficient.
(f) Not well; use a longer wire or reduce potential gradient.
Answer: (a) ≈1.247 V, (b) limits current, (c) no, (d) no, (e) yes, (f) modify by longer wire.
(b) Limits current from standard cell, protects it.
(c) No, because it is shorted near balance.
(d) No, potentiometer is independent of driver cell internal resistance if current constant.
(e) Yes, if potential gradient still sufficient.
(f) Not well; use a longer wire or reduce potential gradient.
Answer: (a) ≈1.247 V, (b) limits current, (c) no, (d) no, (e) yes, (f) modify by longer wire.
Question 3.23
Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor \( R = 10.0 \, \Omega \) is found to be 58.3 cm, while that with the unknown resistance \( X \) is 68.5 cm. Determine the value of \( X \). What might you do if you failed to find a balance point with the given cell of emf \(\varepsilon\)?
Answer & Explanation:
Using balance condition: \( \frac{X}{R} = \frac{l_X}{l_R} \)
\( X = 10.0 \times \frac{68.5}{58.3} \approx 11.75 \, \Omega \).
If no balance, increase potentiometer wire length or reduce current.
Answer: ≈11.75 Ω.
\( X = 10.0 \times \frac{68.5}{58.3} \approx 11.75 \, \Omega \).
If no balance, increase potentiometer wire length or reduce current.
Answer: ≈11.75 Ω.
Question 3.24
Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer & Explanation:
Formula: \( r = R \left( \frac{l_1}{l_2} - 1 \right) \)
\( r = 9.5 \left( \frac{76.3}{64.8} - 1 \right) = 9.5 \times 0.177 \approx 1.68 \, \Omega \).
Answer: ≈1.68 Ω.
\( r = 9.5 \left( \frac{76.3}{64.8} - 1 \right) = 9.5 \times 0.177 \approx 1.68 \, \Omega \).
Answer: ≈1.68 Ω.
📘 Exam Preparation Tip:
These exercise questions will help you understand DC circuit analysis and electrical measurements. You'll learn to apply Ohm's law to various materials, use Kirchhoff's laws to solve complex circuits, calculate equivalent resistance in networks, analyze Wheatstone bridge and potentiometer circuits, and understand meter bridge and other measuring instruments. Essential for circuit design and electrical engineering fundamentals.