Resistance of a System of Resistors – Long Answer Questions
Medium Level (Application & Explanation)
Q1. Why do we combine resistors in circuits, and how does it help in practical situations?
Answer:
We combine resistors to create a desired resistance value when a single resistor of that value is not available. By using series or parallel connections, we can tune the total resistance to almost any required value.
In series, resistances add up, so we can increase resistance; for example, 2 Ω + 3 Ω = 5 Ω. This helps in limiting current safely.
In parallel, the total resistance decreases and is always less than the smallest resistance; two 4 Ω resistors in parallel give 2 Ω.
This is useful in repairs, toys, and DIY circuits, where exact components may be unavailable.
Combining resistors also allows us to design voltage dividers and current limiters, protecting bulbs, LEDs, and sensitive sensors.
Thus, combining resistors gives flexibility, safety, and precision in circuit design.
Q2. Explain the series combination of resistors with a calculation showing current and voltage distribution.
Answer:
In a series combination, the same current flows through each resistor and the total resistance is the sum: R_eq = R1 + R2 + R3.
Example: Resistors of 2 Ω, 3 Ω, and 5 Ω in series with a 12 V battery.
R_eq = 2 + 3 + 5 = 10 Ω.
Using Ohm’s law (V = IR): I = V/R_eq = 12/10 = 1.2 A. This same current flows through all resistors.
The voltage divides based on resistance:
V_2Ω = I × R = 1.2 × 2 = 2.4 V
V_3Ω = 1.2 × 3 = 3.6 V
V_5Ω = 1.2 × 5 = 6.0 V
The sum of the drops equals the supply: 2.4 + 3.6 + 6.0 = 12 V.
Thus, series resistors share the voltage, while the current remains same throughout the chain.
Q3. Describe parallel combination of resistors and show how current divides among branches with a worked example.
Answer:
In a parallel combination, the voltage across each resistor is the same, but the current divides among branches. The equivalent resistance is found using 1/R_eq = 1/R1 + 1/R2 + 1/R3.
Example: 2 Ω, 3 Ω, and 6 Ω across 6 V.
1/R_eq = 1/2 + 1/3 + 1/6 = 1 → R_eq = 1 Ω
Total current: I_total = V/R_eq = 6/1 = 6 A
Branch currents (I = V/R):
Through 2 Ω: I_2Ω = 6/2 = 3 A
Through 3 Ω: I_3Ω = 6/3 = 2 A
Through 6 Ω: I_6Ω = 6/6 = 1 A
Note that smaller resistance draws more current.
Parallel circuits are preferred when we need the same voltage for all devices and independent operation of each branch.
Q4. How would you verify experimentally that total resistance increases in series using ammeter and voltmeter readings?
Answer:
Set up a circuit with a battery (say 6 V), an ammeter in series, and three resistors (2 Ω, 3 Ω, 5 Ω) connected end-to-end. Connect a voltmeter across the battery.
Record the current I using the ammeter.
Theoretical R_eq (series) = 2 + 3 + 5 = 10 Ω.
Using measured values, compute experimental R_exp = V/I. With V = 6 V, if I ≈ 0.6 A, then R_exp ≈ 10 Ω, matching theory.
Check individual voltage drops by placing the voltmeter across each resistor and verify V_i = I × R_i.
Ensure tight connections, minimal contact resistance, and correct meter polarities to reduce errors.
This confirms that in series, total resistance increases, current is same, and voltage divides according to resistor values.
Q5. When should we prefer series vs. parallel connections in daily-life circuits? Provide reasons with examples.
Answer:
Use series when you want to increase resistance or create a voltage drop across parts, such as making a current limiter for LEDs or building a voltage divider. Old decorative strings used series, but one failure caused the entire chain to go off.
Use parallel when devices must receive the same voltage and work independently. Home appliances and wall sockets are in parallel so each gets 220 V and one failure doesn’t stop others.
In series, brightness may differ because voltage divides; in parallel, brightness is consistent at a fixed supply voltage.
Parallel circuits also allow individual switches, whereas series circuits do not.
Therefore, choose series for controlling current/voltage, and parallel for reliable, independent operation at constant voltage.
High Complexity (Analytical & Scenario-Based)
Q6. Design a 3 Ω equivalent resistance using only 2 Ω and 6 Ω resistors. Show at least two valid configurations with reasoning.
Answer:
Goal: R_eq = 3 Ω using only 2 Ω and 6 Ω.
Option 1 (Simple): Two 6 Ω in parallel:
1/R_eq = 1/6 + 1/6 = 2/6 → R_eq = 3 Ω. This is compact and efficient.
Option 2 (Composite): Make 1.5 Ω blocks, then series:
First, one 2 Ω in parallel with one 6 Ω:
1/R = 1/2 + 1/6 = 2/3 → R = 1.5 Ω
Now take two such 1.5 Ω blocks in series:
R_eq = 1.5 + 1.5 = 3 Ω
Reasoning:
Parallel reduces resistance; series adds resistance. By mixing both, we can tune to a target value.
Always verify using formulas: R_series = sum; 1/R_parallel = sum of reciprocals.
Both methods give a precise 3 Ω using allowed values.
Q7. You have a 6 V battery and two identical bulbs rated 6 V, 3 W. Should they be connected in series or in parallel? Justify with calculations of current and power.
Answer:
Each bulb is rated 6 V, 3 W, so it needs the full 6 V. Therefore, connect them in parallel so each gets 6 V.
Bulb resistance: R = V²/P = 36/3 = 12 Ω.
In parallel at 6 V:
Current per bulb: I = P/V = 3/6 = 0.5 A
Total current: I_total = 0.5 + 0.5 = 1.0 A
Each bulb operates at rated brightness (3 W).
If connected in series across 6 V:
Total resistance = 12 + 12 = 24 Ω
Current: I = V/R = 6/24 = 0.25 A
Power per bulb: P = I²R = (0.25)² × 12 = 0.75 W (dim)
Conclusion: Parallel ensures correct voltage and full brightness. Series would under-volt each bulb and make them dim.
Q8. In a parallel network of 2 Ω, 3 Ω, and 6 Ω connected to 12 V, one resistor burns out (open circuit). Predict the changes in total current and brightness. How can you diagnose which branch failed?
Answer:
Initially:
1/R_eq = 1/2 + 1/3 + 1/6 = 1 → R_eq = 1 Ω
I_total = V/R_eq = 12/1 = 12 A
Branch currents: 2 Ω → 6 A; 3 Ω → 4 A; 6 Ω → 2 A (brightness ∝ power P = V²/R).
A large drop to ~6 A suggests 2 Ω failed; a moderate drop to ~8 A suggests 3 Ω failed.
Brightness of remaining branches stays the same because voltage remains 12 V across each.
Q9. You need 3 V from a 9 V source to feed a sensor that draws negligible current. Design a voltage divider and explain how loading affects the output.
Answer:
Use a voltage divider with resistors R1 (top) and R2 (bottom). Output across R2:
V_out = (R2 / (R1 + R2)) × V_in = 3 V with V_in = 9 V.
If the sensor draws negligible current, V_out ≈ 3 V.
With loading (sensor has resistance R_L), R2 becomes parallel to R_L, reducing the effective lower resistance, so V_out drops below 3 V.
To minimize loading error, ensure R_L >> R2 (at least 10×). For R2 = 1 kΩ, choose R_L ≥ 10 kΩ.
Note: Dividers are good for signal-level references, not for power delivery. For heavier loads, use a regulator or a buffer after the divider.
Q10. Explain why household wiring uses parallel connections. Show, with a numerical example, how adding appliances can cause overloading.
Answer:
Homes use parallel wiring so each appliance receives the same 220 V, can be switched independently, and one failure does not stop others. Also, each device operates at its rated voltage, giving proper performance.
Example (approximate resistances using R = V²/P at 220 V):
Heater 1100 W → R ≈ 220²/1100 ≈ 44 Ω
Bulb 100 W → R ≈ 220²/100 ≈ 484 Ω
With both ON in parallel:
Currents: I_heater = 220/44 = 5 A, I_bulb = 220/484 ≈ 0.45 A
Total I ≈ 5.45 A
Add another 1100 W heater:
New I_total ≈ 5 + 5 + 0.45 = 10.45 A
If the circuit is protected by a 10 A MCB/fuse, this can cause overloading, trip the MCB, or heat wires.
Conclusion: In parallel, adding appliances decreases equivalent resistance, increases total current, and may require proper rating of wires and protective devices.
