Q1: A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is (a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer: (d) 25
If the wire is cut into 5 equal parts, each part will have resistance R/5. When connected in parallel:

$\frac{1}{R’} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} = 5 \times \frac{5}{R} = \frac{25}{R} \Rightarrow R’ = \frac{R}{25}$

Hence,

$R/R’ = 25$

.

Q2: Which of the following terms does not represent electrical power in a circuit? (a) I2R
(b) IR2
(c) VI
(d) V2/R

Answer: (b) IR²
Correct formulae for power:

$P = I^2R,\ P = VI,\ P = V^2/R$

.

$IR^2$

is not a formula for power.

Q3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be (a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer: (d) 25 W

$P = V^2 / R$

. Resistance R =

$220^2 / 100 = 484 \Omega$

Now

$P = 110^2 / 484 = 12100 / 484 = 25 \text{ W}$

.

Q4: Ratio of heat produced in series and parallel combinations of equal wires (a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: (d) 4:1
In series: higher equivalent resistance → more heat.
Using

$H \propto R$

and

$R_{series} = 4R,\ R_{parallel} = R/4$

$\frac{H_{series}}{H_{parallel}} = \frac{4R}{R/4} = 16$

, but since potential is constant, it comes out to 4:1.

Q5: How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer: A voltmeter is always connected in parallel with the component across which the potential difference is to be measured.

Q6: Length of a wire and change in resistance when diameter is doubled?
Answer:
Given:

$r = 0.25 \text{ mm} = 0.25 \times 10^{-3} \text{ m},\ \rho = 1.6 \times 10^{-8} \Omega \text{m},\ R = 10 \Omega$

$R = \frac{\rho l}{\pi r^2} \Rightarrow l = \frac{R \pi r^2}{\rho} = \frac{10 \times 3.14 \times (0.25 \times 10^{-3})^2}{1.6 \times 10^{-8}} \approx 1.23 \text{ m}$

If diameter is doubled, new radius

$r’ = 0.5 \text{ mm}$

. New resistance:

$R’ = \frac{\rho l}{\pi r’^2} = R/4 = 2.5 \Omega$

Q7: Plot graph between V and I and calculate resistance

Answer:
Resistance from slope

$R = V/I$

, using any point:

$R = 1.6/0.5 = 3.2 \Omega$

. All points give approx. same.

Q8: 12 V battery, 2.5 mA current. Find resistance.
Answer: Here, V = 12 V and I = 2.5 mA = 2.5 x 10^-3 A

$R = V/I = 12 / (2.5 \times 10^{-3}) = 4800 \Omega$

Q9: A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?
Answer:
Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = VR = 12V13.4Ω = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

Q10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? 
Answer:
Let n resistors of 176 Ω are connected in parallel.

Thus 4 resistors are needed to be connect.

$1/R = n/176 \Rightarrow n = 176/44 = 4$

Q11: Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω
Answer:
Here, R₁ = R₂ = R₃ = 6 Ω.

(i) When we connect R₁ in series with the parallel combination of R₂ and R₃ as shown in Fig. (a).
The equivalent resistance is

(ii) When we connect a series combination of R₁ and R₂ in parallel with R₃, as shown in Fig. (b), the equivalent resistance is

$1/R = 1/12 + 1/6 = 1/4 \Rightarrow R = 4 \Omega$

Q12: Bulbs rated 220 V, 10 W. Max current 5 A.
Answer: Given, current, I = 5 A, voltage, V = 220 V

$P = VI = 220 \times 5 = 1100 \text{ W}$

Number of lamps: =Maximum Power/ Power of Bulb =

$1100\mathrm{/}10=110$

$∴ 110 lamps can be connected in parallel.∴ 110 lamps can be connected in parallel.∴ 110 lamps can be connected in parallel.1100 / 10 = 110$

Q13: Two coils of 24 \Omega: used (i) separately (ii) in series (iii) in parallel.
Answer: (i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V

(ii) When the two coils are connected in series,

(iii) When the two coils are connected in parallel.

$R = 12 \Omega, I = 220/12 \approx 18.33 \text{ A}$

Q14: Compare the power used in the 2 Ω resistor in each of the following circuits
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:
(i) The circuit diagram is shown in figure.
Total resistance, R = 1Ω + 2Ω = 3Ω
Potential difference, V = 6 V

Power used in 2Ω resistor = I²R = (2)² x 2 = 8 W

(ii) The circuit diagram for this case is shown :
Power used in 2 resistor = v2R =422 = 8 W.

[ ∵ Current is different for different resistors in parallel combination.

Q15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ?
Answer:
Power of first lamp (P₁) = 100 W
Potential difference (V) = 220 V

$= (100 + 60)/220 = 160/220 \approx 0.727 \text{ A}$

Q16: Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?

Answer:
Energy used by the TV set = 250 W × 1 hour = 250 Wh

Energy used by the toaster = 1200 W × (10/60) hour
= 1200 × (1/6) = 200 Wh

Conclusion:
The 250 W TV set used for 1 hour consumes more energy (250 Wh) than the 1200 W toaster used for 10 minutes (200 Wh).

Q17: An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:
Given:

$R = 8\, \Omega$

R=8Ω

$I = 15\, \text{A}$

I=15A

$t = 2\, \text{h}$

t=2h (Time is not needed here because we are asked for the rate, i.e., power)

The rate at which heat is developed in the heater is equal to the power consumed.

Using the formula:

$$P = I^2 R$$

P=I2R

$$P = (15)^2 \times 8 = 225 \times 8 = 1800 \, \text{W}$$

P = (15)2×8 = 225×8 = 1800W The rate at which heat is developed in the heater is 1800 watts (W) or 1800 joules per second (J/s).

$P = I^2R = 15^2 × 8 = 1800 \text{ W}$

Q18: Explain the following:

(i) Why is tungsten used almost exclusively for filament of electric lamps?
Answer:
Tungsten is used almost exclusively for the filament of electric lamps because it has a very high melting point (around 3300°C). When electricity flows through the filament, it gets heated up to nearly 2700°C and glows brightly, emitting light without melting. Its durability at high temperature makes it ideal for this purpose.

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Answer:
Alloys are used instead of pure metals in heating devices because:

• Alloys have higher resistivity than pure metals, which helps in producing more heat.
• They do not oxidise or burn easily even at high temperatures.
  Thus, they are safer and more efficient for heating applications.

(iii) Why is the series arrangement not used for domestic circuits?
Answer:
Series arrangement is avoided in domestic circuits because:

• If one appliance fails or gets switched off, the entire circuit breaks, and all appliances stop working.
• It does not allow appliances to be operated independently.
  To avoid inconvenience and allow individual control, parallel arrangement is used in homes.

(iv) How does the resistance of a wire vary with its area of cross-section?
Answer:
The resistance of a wire is inversely proportional to its area of cross-section.
That is,

$R \propto \frac{1}{A}$

R∝A1​
So, if the area increases, resistance decreases, and vice versa. A thicker wire offers less resistance because it allows more electrons to flow through it.

(v) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
Copper and aluminium are commonly used for electricity transmission because:

• They have very low electrical resistances, which reduces power loss.
• They are good conductors of electricity.
• Aluminium is also lighter and more cost-effective than copper.
  These properties make them ideal for efficient and safe electricity transmission.