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Kirchhoff's Junction Law

Potential dif...

Question

Potential difference $V_{A} - V_{B}$ in the network shown is?

1

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Solution

The correct option is A $1$ V

Since , The resistance between two different branches that is $P A Q$ and $P B Q$ is equal that is $5 Ω$ .

Hence, current will be equally distributed in two branches.

$⟹ I_{1} = I_{2} = 1 A$

Now, $V_{P} - V_{A} = I_{1} .2 Ω = 2 V$

Similarly $V_{P} - V_{B} = I_{2} .3 Ω = 3 V$

$V_{A} - V_{B} = (V_{P} - V_{B}) - (V_{P} - V_{A})$

$⟹ V_{A} - V_{B} = 1 V$

Answer-(A)

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Potential difference VA−VB in the network shown is?

The correct option is A 1 VSince , The resistance between two different branches that is PAQ and PBQ is equal that is 5Ω.Hence, current will be equally distributed in two branches.⟹I1=I2=1ANow, VP−VA=I1.2Ω=2VSimilarly VP−VB=I2.3Ω=3VVA−VB=(VP−VB)−(VP−VA)⟹VA−VB=1VAnswer-(A)

Potential difference $V_{A} - V_{B}$ in the network shown is?