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  • A battery of emf  is connected to a group of resistances as shown in figure. The pote

A battery of emf \mathrm{ 10 V} is connected to a group of resistances as shown in figure. The potential difference \mathrm{V_A-V_B} between the points \mathrm{A \: and\: B} is

Option: 1

-2\mathrm{V}


Option: 2

2\mathrm{V}


Option: 3

5\mathrm{V}


Option: 4

\frac{20}{11}\mathrm{V}  


Answers (1)

best_answer

Resistance between upper branch and lower branch in parallel part is same, so equal amount of current flows thgough them.
Let main current is \mathrm{i}.

\mathrm{\therefore i=\frac{V}{R_{eq}}}

Equivalent resistance of circuit,\mathrm{R_{eq}=3+2=5\Omega\Rightarrow\mathrm{i}=\frac{10}{5}=2A }

So, current in each branch \mathrm{=1A}

Now, \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=1 \times 1=1 \mathrm{V}                        (i)

also, \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=1 \times 3=3 \mathrm{V}                        (ii)

Solving eqs. (i) and (ii), we have

\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=3-1=2 \mathrm{~V}

Hence option 2 is correct.

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