NTSE Stage 2 SAT 2016 Question Paper

Question : 30 of 100

Marks: +1, -0

In the circuit diagram shown below,

\V_\A

and

\V_\B

are the potentials at points A and B respectively. Then,

\V_\A – \V_\B

Solution:

From the given circuit diagram

\R_{\eff}={30×15}/{3×15}={30×15}/45=10\Ω

\;\;\;\;\i=3\A

In branch CA current = 1A
In branch CB current = 2A
∴

\;\;\;\;\V_C-\V_\A=10\V

...(i)
and

\V_C-\V_\B=20\V

...(ii)
Solving equation (i) and (ii) we get

\;\;\;\;\;\;\;\;\V_\A-\V_\B=10\V

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