3. Current Electricity
Complete solutions to the Exercises. Maharashtra State Board (Balbharati), Standard Nine Science and Technology.
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Exercises
(A) By which method are the appliances connected?
The appliances are connected in parallel. Each appliance is connected separately between the live wire and the neutral wire through its own socket.
(B) What must be the potential difference across individual appliances?
In a parallel connection the potential difference across every appliance is the same, and equal to the supply voltage. In India this is about 220 V across each appliance.
(C) Will the current passing through each appliance be the same? Justify your answer.
No. The potential difference across each appliance is the same (V), but the appliances have different resistances. By Ohm’s law, the current through each is I = V / R, so an appliance with a smaller resistance draws a larger current. A heater, with low resistance, draws much more current than a table lamp or a TV. The total current from the mains is the sum of the currents through all the appliances.
(D) Why are the domestic appliances connected in this way?
- Every appliance gets the full supply voltage of 220 V, which it is designed to work on.
- Each appliance can be switched on or off independently, without affecting the others.
- If one appliance stops working or its fuse blows, the others keep working, because each has its own separate path for current.
- Each appliance draws only the current it needs, according to its own resistance.
In a series connection none of these would be possible — the voltage would be shared among the appliances, and one fault would switch everything off.
(E) If the T.V. stops working, will the other appliances also stop working? Explain your answer.
No. The appliances are connected in parallel, so each has a separate path between the live and neutral wires. If the TV stops working, only its own path is broken. Current continues to flow through the other paths, and the lamps and the heater keep working normally.
The six symbols are: a resistor, an electric cell, an ammeter, a voltmeter, a closed plug key and a variable resistance (rheostat). The raised part of the outline, above one section of the top wire, is where the voltmeter goes, across the resistor. The completed circuit:
Resistor, ammeter, cell, key and rheostat in series; voltmeter in parallel with the resistor.
- Cell, plug key, rheostat, resistor and ammeter are all connected in series, so the same current flows through each of them. The ammeter measures this current.
- Voltmeter is connected in parallel across the two ends of the resistor (X and Y), so it measures the potential difference across the resistor.
- The positive (+) terminals of the ammeter and voltmeter are connected to the side of the circuit that leads to the positive terminal of the cell.
- The rheostat is used to change the current in the circuit in steps.
Law proved: this circuit is used to verify Ohm’s law. For different settings of the rheostat, the readings of the ammeter (I) and the voltmeter (V) are noted. The ratio V / I comes out the same every time, showing that the current through the resistor is directly proportional to the potential difference across it (at constant physical state).
(A) Which method should he use to connect the bulbs?
He should connect the bulbs in series. When one bulb is connected alone, its resistance is too small, so too large a current flows and the filament burns. In series the effective resistance increases, the current in the circuit decreases, and the supply voltage is shared between the two bulbs.
(B) What are the characteristics of this way of connecting the bulbs depending on the answer of question A above?
- The same current flows through both bulbs.
- The effective resistance is the sum of the individual resistances, so it is larger than either of them.
- The potential difference across the whole arrangement equals the sum of the potential differences across the two bulbs. It is divided in the ratio of their resistances, 15 : 30 = 1 : 2, so the 30 Ω bulb gets twice the potential difference of the 15 Ω bulb.
- Because the current is reduced, both bulbs glow less brightly than when connected singly, and their filaments are protected.
- If one bulb fuses, the circuit breaks and the other bulb also goes off.
(C) What will be the effective resistance in the above circuit?
Answer: the effective resistance of the circuit is 45 Ω.
In the textbook, part C of this question is printed at the top of the next column, just above the picture of the room. It belongs to Question 3, not to Question 1.
(a) Find the average resistance.
Reading 1: R1 = 4 ÷ 9 = 0.444 Ω
Reading 2: R2 = 5 ÷ 11.25 = 0.444 Ω
Reading 3: R3 = 6 ÷ 13.5 = 0.444 Ω
Average resistance = (0.444 + 0.444 + 0.444) ÷ 3 = 0.444 Ω (= 4/9 Ω)
Answer: the average resistance is 0.44 Ω (exactly 4/9 Ω).
(b) What will be the nature of the graph between the current and potential difference?
V / I has the same value for every reading, so V is directly proportional to I. The graph of current against potential difference will be a straight line passing through the origin.
(c) Which law will the graph prove? Explain the law.
The graph proves Ohm’s law.
Ohm’s law: if the physical state of a conductor (its length, area of cross-section, temperature and material) remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
I ∝ V, so I = kV. Writing 1/k = R, the resistance of the conductor, we get V = IR, or R = V / I.
In the table, V / I is the same (0.444 Ω) for all three readings, so the conductor obeys Ohm’s law.
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| ‘A’ Group | ‘B’ Group |
|---|---|
| 1. Free electrons | c. Weakly attached |
| 2. Current | a. V / R |
| 3. Resistivity | d. VA / LI |
| 4. Resistances in series | b. Increases the resistance in the circuit |
Why resistivity = VA/LI: ρ = RA / L, and by Ohm’s law R = V / I. Substituting, ρ = (V / I) × A / L = VA / LI.
Here R = r, L = x, A = a
∴ r = ρ x / a
∴ ρ = r a / x
Unit: ρ = (Ω × m²) ÷ m = Ω m (ohm metre)
Answer: the resistivity is ρ = ra / x and its SI unit is the ohm metre (Ω m).
How the circuit is arranged (from the figure): R3 is in the main line next to the cell, so all the current passes through it. R1 (with the ammeter) lies between C and D; R2 lies between B and E, and key S1 joins E to D — so when S1 is closed, R2 is in parallel with R1. R4 lies between H and I, and key S2 joins F directly to G — so when S2 is closed, it provides a zero-resistance path around R4.
(a) Both S1 and S2 are closed.
- S1 closed: R1 and R2 are in parallel between the left side and D.
- S2 closed: current takes the zero-resistance path F–G through the key, so R4 is short-circuited and no current flows through R4.
- Current flows through R3, divides between R1 and R2, joins again at D and returns to the cell through S2.
- Effective resistance = R3 + R1R2 / (R1 + R2). This is the smallest of the three cases, so the current in the circuit is the largest. The ammeter shows only the part of the current that passes through R1.
(b) Both S1 and S2 are open.
- S1 open: the branch containing R2 is broken, so no current flows through R2.
- S2 open: current can no longer go through the key and must pass through R4.
- R3, R1 and R4 are now in series, and the same current flows through all of them and through the ammeter.
- Effective resistance = R1 + R3 + R4. This is the largest of the three cases, so the current in the circuit is the smallest.
(c) S1 is closed but S2 is open.
- S1 closed: R1 and R2 are in parallel.
- S2 open: current flows through R4.
- The parallel pair (R1, R2) is in series with R3 and R4. Current flows through all four resistors.
- Effective resistance = R3 + R4 + R1R2 / (R1 + R2). The current is less than in case (a) but more than in case (b).
In short: current in case (a) > case (c) > case (b). Closing S1 adds a parallel path (resistance falls); closing S2 bypasses R4 (resistance falls further).
(a) Current I flows through x1, x2 and x3 — series. The same current flows through every resistor only in series.
(b) x is larger than x1, x2 and x3 — series. In series the effective resistance is the sum, so it exceeds each one.
(c) x is smaller than x1, x2 and x3 — parallel. In parallel the effective resistance is less than the least individual resistance.
(d) The potential difference across x1, x2 and x3 is the same — parallel. All resistors in parallel are connected between the same two points.
(e) x = x1 + x2 + x3 — series. This is the formula for effective resistance in series.
(f) x = 1 / (1/x1 + 1/x2 + 1/x3) — parallel. This comes from 1/x = 1/x1 + 1/x2 + 1/x3, the formula for parallel connection.
(A) The resistance of a 1 m long nichrome wire is 6 Ω. If we reduce the length of the wire to 70 cm, what will its resistance be?
Same wire (same material and area), so R ∝ L
R2 / R1 = L2 / L1
R2 = 6 × 0.7 ÷ 1 = 4.2 Ω
Answer: the resistance of the 70 cm wire will be 4.2 Ω.
(B) When two resistors are connected in series, their effective resistance is 80 Ω. When they are connected in parallel, their effective resistance is 20 Ω. What are the values of the two resistances?
Parallel: R1R2 / (R1 + R2) = 20
∴ R1R2 = 20 × 80 = 1600 …… (2)
(R1 − R2)² = (R1 + R2)² − 4R1R2 = 6400 − 6400 = 0
∴ R1 = R2, and from (1), R1 = R2 = 40 Ω
Check: series 40 + 40 = 80 Ω; parallel (40 × 40) ÷ 80 = 20 Ω ✓
Answer: both resistances are 40 Ω each.
(C) If a charge of 420 C flows through a conducting wire in 5 minutes, what is the value of the current?
I = Q / t = 420 ÷ 300 = 1.4 A
Answer: the current is 1.4 A.
Points to ask the electrician about and note down: