Solved 2024 Sample Question Paper CBSE Class 10 Science — Question 3

(i) (a) When switch A is connected to position 2, circuit A-2 is complete and only 12 Ω lamps are on.

(b) When switch A is connected to position 3, circuit A-3 is complete and only 4 Ω lamps are on.

(ii) We know that when the resistances are connected in parallel, they have the same potential difference across their ends, hence, potential difference across each lamp when lit will be 12 V.

(iii) When the wire is connected to position 2, 12 Ω lamps are on.

Resistance of each lamp = 12 Ω

Voltage across each lamp = 12 V.

Using Ohm's Law: V = IR

Substituting we get,

12 = I x 12

I = $\dfrac{12}{12}$ = 1 A

Hence, current across each 12 Ω lamp is 1 A.

When the wire is connected to position 3, 4 Ω lamps are on.

Resistance of each lamp = 4 Ω

Voltage across each lamp = 12 V.

Using Ohm's Law: V = IR

Substituting we get,

12 = I x 4

I = $\dfrac{12}{4}$ = 3 A

Hence, current across each 4 Ω lamp is 3 A.

OR

(iv) Given,

All lamps are in parallel so V across each lamp = 12 V

We know, P = $\dfrac{V^2}{R}$

Substituting we get,

For 12 Ω lamps:

P = $\dfrac{12 \times 12}{12}$ = 12 W

For 4 Ω lamps:

P = $\dfrac{12 \times 12}{4}$ = 36 W

Hence, power across 4 Ω lamps will be higher.