Resistance of a System of Resistors – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Explain, with examples, how the total resistance changes when resistors are connected in series. Why does this happen?

Answer:

When resistors are connected in series, the total resistance increases.
This is because the current has to pass through each resistor one after the other, facing the opposition of every resistor in the path.
The formula used is $R_{total} = R_1 + R_2 + R_3 + ...$ .
For example, if three resistors of 2 Ω, 3 Ω, and 5 Ω are in series, total resistance is $2 + 3 + 5 = 10$ Ω.
The increased resistance means the circuit will allow less current to flow for a given voltage.
This is like water flowing through a series of narrow pipes; each pipe adds more resistance to the flow.

Q2. What is meant by a parallel combination of resistors? How does it affect the total resistance? Illustrate with a simple calculation.

Answer:

In a parallel combination, resistors are connected so that their ends are joined at the same two points.
The voltage across each resistor is the same.
The total resistance decreases and is less than the smallest individual resistor.
The formula is $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + ...$ .
For resistors of 4 Ω and 12 Ω in parallel: $\frac{1}{R} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$ , so $R = 3$ Ω.
More paths for current means easier flow and smaller total resistance.

Q3. Give two everyday examples each where resistors are connected (a) in series and (b) in parallel, explaining why that particular arrangement is used.

Answer:

Series:
- (1) Some decorative lights, where bulbs are connected one after another; if one bulb fails, all go off, showing series connection.
- (2) Old model torches or flashlights, where cells’ resistances add up to increase resistance and reduce current drain.
Parallel:
- (1) Household wiring: sockets and appliances are connected in parallel so each gets full voltage and works independently.
- (2) Fuses: Branches of electrical circuits may have fuses in parallel to ensure each branch is protected by a separate fuse.
The arrangement is chosen based on desired current flow and independence of components.

Q4. Why is parallel arrangement preferred for domestic electrical circuits? Explain in detail.

Answer:

Parallel arrangement ensures each appliance gets the same voltage as the source.
Appliances can be operated independently. Switching one off doesn’t affect others.
If one device fails, the overall circuit remains intact; only that branch stops working.
This arrangement also allows different appliances to draw the current they need, as per their resistance.
Also, total resistance decreases, so more overall current can be drawn when needed.
Thus, safety, independence, and
convenience
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make parallel arrangement ideal for homes.

Q5. How do current and potential difference behave in series and parallel resistor arrangements? Explain with suitable examples.

Answer:

In a series circuit, the current is the same through all resistors, but the potential difference (voltage) is divided among them based on their resistance.
For example, in a series of three bulbs with a 12 V battery, if they have equal resistance, each will get 4 V.
In a parallel circuit, the potential difference across each resistor is the same as the source, while the current divides among branches depending on resistance.
For example, if a 12 V battery is connected to two parallel resistors of 3 Ω and 6 Ω, both get 12 V, but the 3 Ω resistor allows twice as much current as the 6 Ω one.
Thus, series: current same, voltage divides; parallel: voltage same, current divides.

High Complexity (Analysis & Scenario-Based)

Q6. If three resistors of values 2 Ω, 3 Ω, and 6 Ω are available, how would you connect them to get maximum and minimum total resistance? Calculate both values.

Answer:

Maximum resistance is obtained by connecting all resistors in series:
$R_{max} = 2 + 3 + 6 = 11$ Ω.
Minimum resistance is obtained by connecting all in parallel:
$\frac{1}{R_{min}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Rightarrow R_{min} = 1$ Ω.
This is because series adds resistances, giving higher value, while parallel reciprocals sum up, giving lowest possible resistance (less than the smallest one).
Choice of arrangement depends on whether high or low resistance is needed in the circuit.
This flexibility is essential in designing different electronic circuits.
Always check if the arrangement meets the needs of voltage and current in the application.

Q7. A student sets up a circuit with a battery and two bulbs in series. If one bulb fuses, what happens to the other bulb? What if they were in parallel? Explain why.

Answer:

In series connection:
- If one bulb fuses, the circuit becomes open (broken).
- The current stops flowing, so the other bulb also goes off.
In parallel connection:
- The bulbs are connected across the same two points.
- If one bulb fuses, the other bulb remains lit because its own path for current is still complete.
This happens because current in series has only one path; breaking it stops all flow.
In parallel, each component has its own path; breaking one does not affect the others.
That is why parallel wiring is preferred in homes and appliances.

Q8. Suppose you have three resistors, each of resistance 6 Ω. Show two different ways to connect them to get 4 Ω, and explain the reasoning with calculations.

Answer:

Way 1 (Series-Parallel Combo):
- Connect two resistors in parallel: $\frac{1}{R_1} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \Rightarrow R_1 = 3$ Ω.
- Now, connect the third resistor in series with this combination:
  Total $R = R_1 + 6 = 3 + 6 = 9$ Ω (But this is not 4 Ω, so try another way).
Way 2 (All in Parallel):
- $\frac{1}{R} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \Rightarrow R = 2$ Ω.
Correct Way (One in Series with Parallel):
- Connect two in parallel and the third in series with the parallel combo:
- Two in parallel: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \Rightarrow R_p = 3$ Ω.
- Series with third: $R_{total} = 6 + 3 = 9$ Ω (already done).
There is actually no way to get exactly 4 Ω with three 6 Ω resistors using only series and parallel combinations.
This shows the limitation of combining fixed resistors. Sometimes, the required value cannot be obtained with a given set.

Q9. Analyze a scenario: Why do decorative lights connected in series go out completely when even a single bulb fuses, whereas those in parallel do not? Discuss the physics behind it with proper explanation.

Answer:

In a series connection, all bulbs are on a single current path.
If one bulb fuses (its filament breaks), the circuit gets broken (open) and current cannot flow at all.
Due to this, all bulbs go out together.
In parallel connection, each bulb has its own path connecting between the two supply points.
If one bulb fuses, only that single branch opens; the rest of the bulbs are still connected and keep glowing.
Thus, the difference is due to the circuit paths: series—one path, parallel—multiple paths for current.

Q10. Suppose you need a net resistance of 8 Ω but you have only two resistors: one of 12 Ω and one of 24 Ω. Show how you can connect them to get as close as possible to 8 Ω, explaining your calculations.

Answer:

Try series connection: $R_{series} = 12 + 24 = 36$ Ω (too high).
Try parallel connection:
$\frac{1}{R_{parallel}} = \frac{1}{12} + \frac{1}{24} = \frac{2 + 1}{24} = \frac{3}{24} = \frac{1}{8} \Rightarrow R_{parallel} = 8$ Ω.
So, connecting the 12 Ω and 24 Ω resistors in parallel will give exactly 8 Ω.
This works because the parallel arrangement allows the current to split between two paths, lowering the total opposition.
This is a practical use of parallel connection to get the desired resistance when exact values are not available.
Always use parallel formula to check if you can get the required value with given resistors.