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Q.
A battery of emf, $10\, V$ is connected to resistances as shown in figure. The potential difference $V_{A}-V_{B}$ between the points $A$ and $B$ is
Current Electricity
Solution:
Total resistance of circuit $=3+\frac{4 \times 4}{4+4}=3+2=5\, \Omega$
Current in circuit, $I=\frac{10\, V }{5\, \Omega}=2\, A$
Current through arm $C A D$ or $C B D=1\, A$
Thus $V_{C}-V_{A}=1 \times 1=1 V$ and $V_{C}-V_{B}=1 \times 3=3\, V$
Hence $\left.V_{A}-V_{B}=\left(V_{C}-V_{B}\right)-V_{C}-V_{A}\right)=3-1=2\, V$
