CG Board Class 10 Science Chapter 6 Solutions: Electric Current and Circuit (विद्युत धारा एवं परिपथ) (2026-27)

(i) Which one of the options given below is a good conductor of electricity:**

• (a) Wood piece
• (b) Sheet of paper
• (c) Copper wire
• (d) Glass rod

(c) Copper wire

Explanation: Copper is a metal containing free electrons, making it an excellent conductor of electricity. Wood, paper, and glass are insulators.

Answer: (b) Ammeter

Explanation: An ammeter has very low resistance and is connected in series in a circuit to measure electric current.

Answer: (b) A fixed ratio

Explanation: According to Ohm’s law (V = IR), the ratio V/I = R remains constant (fixed ratio) for a conductor at constant temperature.

Answer: (d) 100 A

Explanation: Using Ohm’s law formula I = V/R = (220 V)/(2.2 Ω) = 100 A.

Answer: (b) 20 Ω

Explanation: For two 10 Ω resistors connected in series, equivalent resistance R = R₁ + R₂ = 10 Ω + 10 Ω = 20 Ω.

(i) The potential difference of resistors connected in a series would be …………………….**

(ii) In resistors connected in a parallel combination, the current flowing across each resistor would be …………………………….

(iii) We show the consumption of electricity in kWh or in …………………………..

(iv) The filament used in electric bulbs is made of ……………………. metal.

(v) The ………………………. of a conductor increases with temperature.

• (i) Different (Divided across resistors)
• (ii) Different
• (iii) Units (Board of Trade Units)
• (iv) Tungsten
• (v) Resistivity / Resistance

8.33 Ω

Definition:** A fuse is a safety device used in electrical circuits to protect appliances from damage due to excessive current flow, short circuits, or overloading.

• Working Material: It consists of a thin wire made of a metal or alloy having a low melting point and high resistance, such as an alloy of lead and tin (or aluminium/copper). When current exceeds the safe limit, the fuse wire heats up, melts, and breaks the circuit.

Ohm’s Law:** At a constant temperature and physical state of a conductor, the electric current (I) flowing through the conductor is directly proportional to the potential difference (V) applied across its ends

V ∝ I ⇒ V = R · I or R = V/I

where R is a constant called the resistance of the conductor.

• Graph Verification: If we plot potential difference V on the Y-axis and current I on the X-axis, we obtain a straight line passing through the origin. The constant slope of this line represents the resistance R.

Given:**

• Electrical Energy (W) = 2400 J
• Time (t) = 1 minute = 60 seconds
• Formula:

Electric Power (P) = (Work done / Energy (W))/(Time (t))

• Calculation:

P = (2400 J)/(60 s) = 40 Watts (W)

• Result: The power of the bulb is 40 W.

Given:**

• Power (P) = 3 kW = 3000 W
• Voltage (V) = 220 V
• (a) Electric Current (I):

P = V × I ⇒ I = P/V = (3000)/(220) = 13.63 A

• (b) Resistance of Heater (R):

R = V/I = (220)/(13.636) = 16.13 Ω (or R = V²/P = (220 × 220)/(3000) = 16.13 Ω)

• (c) Electricity Cost for 10 hours:
• Energy consumed = P × t = 3 kW × 10 h = 30 kWh (units)
• Cost = 30 units × Rs. 1.50/unit = Rs. 45

(i) Why are domestic circuits not arranged in a series combination?**

(ii) Why is only tungsten metal used to make filaments of bulbs?

(i) ?

• (i) Domestic circuits are not connected in series because

1. In series, if one appliance stops working or fails, the whole circuit breaks and all other appliances turn off.
2. The supply voltage (220 V) divides among appliances, providing insufficient voltage to each device.

• (ii) Tungsten is exclusively used for bulb filaments because

1. It has an extremely high melting point ( ≈ 3380°C).
2. It has high resistivity and does not oxidize (burn) easily at high glowing temperatures.

• (i)
• (ii)

Given:**

• Current (I) = 0.5 A
• Time (t) = 20 minutes = 20 × 60 s = 1200 seconds
• Formula & Calculation:

Q = I × t = 0.5 A × 1200 s = 600 Coulombs (C)

• Result: The quantity of charge is 600 C.

Daily Energy Consumption:**

1. 4 bulbs of 40 W for 5 h = 4 × 40 × 5 = 800 Wh
2. 2 bulbs of 60 W for 6 h = 2 × 60 × 6 = 720 Wh
3. 3 fans of 80 W for 6 h = 3 × 80 × 6 = 1440 Wh

• Total Daily Energy = 800 + 720 + 1440 = 2960 Wh = 2.96 kWh
• Monthly Energy (30 days):

Total Units = 2.96 kWh × 30 = 88.8 units (kWh)

• Electricity Cost:

Cost = 88.8 units × Rs. 0.50 = Rs. 44.40

(i) What would be the effect on the circuit if one bulb out of three bulbs arranged in a series combination becomes fused?**

(ii) What would be the effect if one bulb out of three arranged in parallel combination malfunctions?

(i) ?

• (i) Series Circuit: The entire circuit breaks because current has only one path. Consequently, the remaining two bulbs will also stop glowing.
• (ii) Parallel Circuit: Current flows through separate independent branches. The remaining two bulbs will continue to glow normally without any change in potential difference or performance.
• (i) :
• (ii) :

Parallel part (4 Ω and 8 Ω):**

1/R_p = 1/4 + 1/8 = 2+1/8 = 3/8 ⇒ R_p = 8/3 = 2.66 Ω

• Series total (2 Ω + 4 Ω + R_p):

R_eq = 2 + 4 + 2.66 = 8.66 Ω

• Result: Equivalent resistance = 8.66 Ω.

In a series circuit, total voltage V = V₁ + V₂.

• V₂ = V – V₁ = 6 V – 2 V = 4 Volts.

Let resistance of each wire be R.

• Series resistance R_s = R + R = 2R.
• Parallel resistance R_p = (R · R)/(R + R) = R/2.
• Heat generated at constant voltage V in time t: H = V²/R t.
• H_s = V²/2R t
• H_p = V²/R/2 t = (2 V²)/(R) t
• Ratio of heat generated (H_s : H_p)

H_s/H_p = (V² / 2R)/(2 V² / R) = 1/4

• Result: The ratio of heat generated is 1 : 4 (or 4 : 1 if parallel to series).

Total series resistance R_s = 2 + 3 + 4 + 5 + 12 = 26 Ω.

• Total circuit current I = V/R_s = (9 V)/(26 Ω) = 0.346 A.
• In a series circuit, current through all resistors is equal. Hence, current across 2 Ω resistor is 0.346 A (or ≈ 0.35 A).

1. Electric Heater / Room Heater: Converts electrical energy into heat using high resistance Nichrome wire coil.
2. Electric Iron: Uses a Nichrome heating element enclosed inside heavy metal plates to produce heat for ironing clothes.
3. Electric Fuse: Safety wire with low melting point that melts during overload to break the circuit.
4. Electric Water Geyser / Electric Kettle: High-power heating element heats water directly.

Resistance (R):**

R = V²/P = (200 × 200)/(100) = 400 Ω

• Energy Consumed:

Energy = 5 bulbs × 100 W × 4 h = 2000 Wh = 2 kWh (units)

• Cost:

Cost = 2 units × Rs. 0.50 = Rs. 1.00

Combination:** Connect the 3 Ω and 5 Ω resistors in parallel with each other, and connect this parallel combination in series with the 2 Ω resistor? Wait, let’s verify exact calculation

• If 3 Ω and 5 Ω in parallel: R_p = (3 × 5)/(3 + 5) = 15/8 = 1.875 Ω.
• Wait! If 3 Ω and 5 Ω in series (8 Ω), connected in parallel with 2 Ω

R_eq = (8 × 2)/(8 + 2) = 16/10 = 1.6 Ω

• Wait, if 2 Ω and 3 Ω in series (5 Ω), connected in parallel with 5 Ω

R_eq = (5 × 5)/(5 + 5) = 25/10 = 2.5 Ω

• Correct Connection: Connect the 2 Ω and 3 Ω resistors in series (giving 5 Ω), and connect this series branch in parallel with the 5 Ω resistor.
• Verification: R_s = 2 + 3 = 5 Ω. Equivalent parallel resistance R_eq = (5 × 5)/(5 + 5) = 2.5 Ω.