The above circuit is a part of an electrical device. Use the

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:

(i) Potential Difference across R₂:

Since Resistances of 4 Ω and R₂ are connected in parallel,

∴ Potential Difference across R₂ = 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 R₂ resistor = 6 Volts

(ii) Value of the resistance R₂:

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 R₂ resistor = 2 – 1.5 = 0.5 Amp

The Potential Difference across R₂ resistor = 6 Volts (calculated in part (i) above)

∴ Resistor R₂ = $\frac{potential~difference}{current} = \frac{6}{0.5}$ = 12 Ω

Therefore, the value of R₂ resistor is 12 Ω.

(iii) Value of resistance R₁:

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 R₂ = 6 Volts (calculated in part (i) above)

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

∴ Battery voltage = PD across R₁ + PD across parallel combination of 4 Ω and R₂ + PD across 2 Ω resistor

∴ 12 = PD across R₁ + 6 + 4

∴ PD across R₁ = 12 – 6 – 4 = 2 Volts

∵ Current in R₁ resistor = 1 Amp (from the diagram)

∴ Resistor R₁ = $\frac{V}{I} = \frac{2}{2}$ = 1 Ω

Therefore, the value of resistance R₁ is 1 Ω.

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