First slide
Sources of emf and Kirchhoff's laws

In the circuit shown:


The current in the circuit I =(2010)(7.5+0.5+1+1)


The potential difference across PS,VPS=7.5×1


The potential difference across Q,R,VQR=1×1


As point G is connected to earth, hence potentials of R and S are zero.

The direction of the current in the circuit is from P to S, hence point P is at higher potential.


 Similarly,  VQ=1V

Hence (b), (c) and (d) are correct.

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