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A current of $\mathrm{2 A}$ flows in the system of conductors as shown in the figure. The potential difference $\mathrm{V_P-V_R}$ will be:

Option: 1
$-2\mathrm{V}$

Option: 2
$\mathrm{-1V}$

Option: 3
$\mathrm{+1V}$

Option: 4
$\mathrm{+2V}$

Answers (1)

best_answer

Resistance between $\mathrm{Q \: and \: S,}$

$\mathrm{R'=\frac{5\times10}{5+10}=\frac{10}{3}\Omega}$

Potential difference across $\mathrm{Q}$ and $\mathrm{S},$

$\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{S}}=\frac{2 \times 10}{3}=\frac{20}{3} \mathrm{~V}$

Current through arm $\mathrm{QPS, I_1=\frac{20}{3 \times 5}=\frac{4}{3} \mathrm{~A}}$

$\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}=\frac{4}{3} \times 2=\frac{8}{3} \mathrm{~V}$

Current through arm $\mathrm{QRS, I_2=\frac{20 / 3}{10}=\frac{2}{3} \mathrm{~A}}$

$\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{R}}=\frac{2}{3} \times 3=2 \mathrm{~V}$

From equations (i) and (ii),

$\mathrm{ V_P-V_R=\left(V_Q-V_R\right)-\left(V_Q-V_P\right) }$

$\mathrm{=2-\frac{8}{3}=\frac{-2}{3} \approx-1 V}$

Hence option 2 is correct.

Posted by

shivangi.bhatnagar

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