In-text Questions — Parallel Circuits — Electricity

1. 13 marksNCERT Cl-10 Science, In-text Qs after §11.6.2, Q1

   Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω and 10⁶ Ω.

   Hint. A huge resistor in parallel with a small one carries almost no current — so it barely changes the combined resistance.

   Step 1 — (a) 1 Ω and 10⁶ Ω. 1/R = 1/1 + 1/10⁶ = 1.000001, so R ≈ 0.999999 Ω — essentially 1 Ω.

   Step 2 — (b) 1 Ω, 10³ Ω and 10⁶ Ω together. 1/R = 1/1 + 1/1000 + 1/10⁶ = 1.001001, so R ≈ 0.999 Ω — again essentially 1 Ω.

   Step 3 — Explain why both cases land near 1 Ω. Since a much larger resistance in parallel with a small one lets almost all the current take the low-resistance path, the huge resistors barely affect the combined value.

   ✦ Answer: In both cases, the equivalent resistance is only slightly less than 1 Ω — the much larger resistors have almost no effect on the combination.

   Where students slip. Assuming adding more large resistors in parallel would noticeably change the result — in parallel, an enormous resistance contributes a vanishingly small share of current, so it barely moves the combined resistance below the smallest resistor present.

2. 23 marksNCERT Cl-10 Science, In-text Qs after §11.6.2, Q2

   An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

   Hint. Find each appliance's current separately first, then add them up.

   Step 1 — Find each appliance's current. I_lamp = 220/100 = 2.2 A; I_toaster = 220/50 = 4.4 A; I_filter = 220/500 = 0.44 A.

   Step 2 — Add them for the total current. I_total = 2.2 + 4.4 + 0.44 = 7.04 A.

   Step 3 — Find the iron's resistance. Since the iron takes this same total current at 220 V, R_iron = V/I = 220/7.04 = 31.25 Ω.

   ✦ Answer: The iron's resistance is 31.25 Ω, and the current through it is 7.04 A.

   Where students slip. Averaging the three resistances instead of adding the three currents — in a parallel circuit, it's the currents (not the resistances) that add directly.

3. 32 marksNCERT Cl-10 Science, In-text Qs after §11.6.2, Q3

   What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

   Hint. Think about what happens to voltage per device, and what happens to the rest of the circuit if one device fails.

   Step 1 — Voltage advantage. Each device gets the full source voltage in parallel, letting devices with different current needs all operate at their rated voltage.

   Step 2 — Independence advantage. If one device fails or is switched off, the others keep working, since each branch is independent.

   Step 3 — Resistance advantage. The total resistance decreases (rather than adding up as in series), allowing the overall current to adjust to what's actually needed.

   ✦ Answer: In parallel, each device gets the full voltage, devices keep working independently of each other, and the combined resistance is lower than any single branch — none of which is true in series.

   Where students slip. Saying parallel circuits 'use less current overall' — total current can actually be higher in parallel (since resistance drops); the real advantages are about voltage per device and independence, not lower overall current draw.

4. 43 marksNCERT Cl-10 Science, In-text Qs after §11.6.2, Q4

   How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

   Hint. Try combining two of the three resistors first, and see what's left to reach the target with the third.

   Step 1 — (a) Target 4 Ω. Put 3 Ω and 6 Ω in parallel: 1/R = 1/3 + 1/6 = 1/2, so R = 2 Ω. Adding the 2 Ω resistor in series with this gives 2 + 2 = 4 Ω.

   Step 2 — (b) Target 1 Ω. Put all three in parallel: 1/R = 1/2 + 1/3 + 1/6 = 6/6 = 1, so R = 1 Ω.

   ✦ Answer: (a) 3 Ω and 6 Ω in parallel, in series with the 2 Ω resistor. (b) All three resistors in parallel.

   Where students slip. Trying to reach 4 Ω by putting all three in series (which gives 11 Ω) — the target values require a mix of parallel and series, not a single simple arrangement.

5. 53 marksNCERT Cl-10 Science, In-text Qs after §11.6.2, Q5

   What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

   Hint. The extremes come from the two simplest possible combinations — no need to try every mixed arrangement.

   Step 1 — (a) Highest resistance. Connecting all four in series gives the maximum: 4 + 8 + 12 + 24 = 48 Ω.

   Step 2 — (b) Lowest resistance. Connecting all four in parallel gives the minimum: 1/R = 1/4 + 1/8 + 1/12 + 1/24 = 6/24 + 3/24 + 2/24 + 1/24 = 12/24 = 1/2, so R = 2 Ω.

   ✦ Answer: Highest = 48 Ω (all in series); lowest = 2 Ω (all in parallel).

   Where students slip. Trying mixed series-parallel arrangements to search for the extremes — the absolute highest and lowest are always given by the simple all-series and all-parallel combinations respectively, not by any mixed arrangement.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc111.pdf) — seven in-text question sets (23 questions total, not 16 as some older manifests claim) plus one end-of-chapter Exercise (18 questions, correctly counted). Unchanged by rationalisation. Table 11.2's resistivity values (used for several answers) were read directly from the book, including nichrome's 100 × 10⁻⁶ Ω·m.. Questions are referenced from the NCERT textbook for identification.