**Q) As shown in the figure above A and B are two lamps. Lamp A is rated at 12 V,24W. Lamp B is rated at 6.0 V. When lamp B operates at its rated voltage, the current in it is 3.0 A. The values of R1 and R2 are chosen so that both lamps operate at their rated voltages.
Based on the information given, answer the following:**

(i) Calculate the current in Lamp A

(ii) State and give reason for the reading of the Voltmeter

(iii) Calculate the resistance of R2

(iv) Find the value of the resistance R1

**(CBSE Sample Paper – 2024-25)**

**Ans: **

**STEP BY STEP SOLUTION**

**(i) Current in Lamp A:**

We know that the power is given by, P = V x I

∴ I =

Given: Power rating of lamp A = 24 W

Voltage rating of lamp A = 12 V

∴ Rated current in lamp A = = 2 Amp

**Therefore, Lamp A will have current of 2 Amp**

**(ii) State and give reason for the reading of the Voltmeter**

Voltmeter is connected parallel to loads and in parallel, potential difference is always same.

∴ reding of voltmeter = potential difference across lamp A = 12 Volts

**Therefore, reading of voltmeter will be 12 volts.**

**(iii) Calculate the resistance of R2:**

Voltage across lamp A = Voltage across (lamp B + Resistance R2)

12 v = 6 V + Voltage across Resistance R2

∴ Voltage across Resistance R2 = 12 – 6 = 6 Volts

Current in Lamp B = 3 Amp

Since Lamp B and Resistance R2 are connected in series,

∴ Current in Resistance R2 = 3 Amp

Resistance R2 =

= = 2 Ω

**Therefore, the value of Resistance R2 is 2 Ω**

**(iv) Find the value of the resistance R1**

Potential Difference across R1 = 15 – 12 = 3 Volts

Current in resistance R1 = Current in Lamp A + Current in Lamp B

= 3 + 2 = 5 Amp

∵ V = I x R

∴ R = = 0.6 Ω

**Therefore Resistance R1 is 0.6 Ω**

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