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Q.
In the circuit shown in figure, the potential difference between the points $A$ and $B$ will be
Current Electricity
Solution:
The circuit diagram may be redrawn as shown here.
Obviosuly, $I_{C A D}=I_{C B D}=\frac{2}{15} A$
$\therefore V_{C}-V_{A}=\frac{2}{15} A \times 5\, \Omega=\frac{2}{3} V$
and $V_{C}-V_{B}=\frac{2}{15} A \times 10\, \Omega=\frac{4}{3} V$
$\therefore V_{A}-V_{B}=\left(V_{C}-V_{B}\right)-\left(V_{C}-V_{A}\right)$
$=\frac{4}{3} V-\frac{2}{3} V=\frac{2}{3} V$
