**Q) The above circuit is a part of an electrical device. Use the information given in the question to calculate the following:
**

**(i) Potential Difference across R2.
**

**(ii) Value of the resistance R2.
**

**(iii) Value of resistance R1**

**(CBSE Sample Paper – 2024-25)**

**Ans: **

**STEP BY STEP SOLUTION**

**(i) Potential Difference across R _{2 }:**

Since Resistances of 4 Ω and R_{2} are connected in parallel,

∴ Potential Difference across R_{2} = Potential Difference across 4 Ω

From the diagram, Current in 4 Ω resistor = 1.5 Amp

∴ The Potential Difference across 4 Ω resistor, V = I x R = 1.5 x 4 = 6 Volts

**Therefore, the Potential Difference across R _{2} resistor = 6 Volts**

**(ii) Value of the resistance R _{2 }:**

From the diagram, total current in the circuit = 2 Amp

and the current in 4 Ω resistor = 1.5 Amp

We know that, in parallel combination, current gets divided

∴ the current in R_{2} resistor = 2 – 1.5 = 0.5 Amp

The Potential Difference across R_{2} resistor = 6 Volts (calculated in part (i) above)

∴ Resistor R_{2} = = 12 Ω

**Therefore, the value of R _{2} resistor is 12 Ω.**

**(iii) Value of resistance R _{1 }:**

From the diagram, we can see that:

a) Battery voltage = 12 Volts

b) Potential Difference across 2 Ω resistor = 2 x 2.0 = 4 Volts

Also Potential Difference across parallel combination of 4 Ω and R_{2} = 6 Volts (calculated in part (i) above)

Next, we know that in Series, potential difference gets distributed

∴ Battery voltage = PD across R_{1} + PD across parallel combination of 4 Ω and R_{2} + PD across 2 Ω resistor

∴ 12 = PD across R_{1} + 6 + 4

∴ PD across R_{1} = 12 – 6 – 4 = 2 Volts

∵ Current in R_{1} resistor = 1 Amp (from the diagram)

∴ Resistor R_{1} = = 1 Ω

**Therefore, the value of resistance R _{1} is 1 Ω.**

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