If in the circuit shown below, the internal resistance of the battery is $1.5$ $\Omega$ and $V_P$ and $V_Q$ are the potentials at $P$ and $Q$ respectively, what is the potential difference between the points $P$ and $Q$
Diffcult
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(d) ${R_{eq}} = \frac{5}{2}\Omega $
$i = \frac{{20}}{{\frac{5}{2} + 1.5}} = 5\,A$
Potential difference between $X$ and $ P$,
${V_X} - {V_P} = \left( {\frac{5}{2}} \right) \times 3 = 7.5V$....$(i)$
${V_X} - {V_Q} = \frac{5}{2} \times 2 = 5\,V$…$(ii)$
On solving $(i)$ and $(ii)$ ${V_P} - {V_Q} = - 2.5{\rm{ }}\,volt\,{\rm{; }}{V_Q} > {V_P}$.
Short Trick : $({V_P} - {V_Q}) = \frac{i}{2}({R_2} - {R_1})$$ = \frac{5}{2}(2 - 3) = - \,2.5$
==> ${V_Q} > {V_P}$
$i = \frac{{20}}{{\frac{5}{2} + 1.5}} = 5\,A$
Potential difference between $X$ and $ P$,
${V_X} - {V_P} = \left( {\frac{5}{2}} \right) \times 3 = 7.5V$....$(i)$
${V_X} - {V_Q} = \frac{5}{2} \times 2 = 5\,V$…$(ii)$
On solving $(i)$ and $(ii)$ ${V_P} - {V_Q} = - 2.5{\rm{ }}\,volt\,{\rm{; }}{V_Q} > {V_P}$.
Short Trick : $({V_P} - {V_Q}) = \frac{i}{2}({R_2} - {R_1})$$ = \frac{5}{2}(2 - 3) = - \,2.5$
==> ${V_Q} > {V_P}$
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