Question

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. the potential difference between the points $P$ and $Q$ is

• A. $zero$
• B. $4V(V_p>V_Q)$
• C. $4V(V_Q>V_P)$
• D. $2.5V(V_Q>V_P)$

Answer 1

Answer: (D)

$2.5V(V_Q>V_P)$

from circuit diagram : $R_{eq}=\frac{5}{2}\Omega$

$i=\frac{20}{\frac{5}{2}+1.5}=5A$

Potential difference between $X$ and $P$,

$V_X-V_P=\left(\frac{5}{2}\right)\times 3=7.5V$ ....(i)

$V_X-V_Q=\frac{5}{2}\times 2=5V$

On solving (i) and (ii) $V_P-V_Q=-2.5volt;V_Q>V_P$

Short trick: $(V_P-V_Q)=\frac{i}{2}(R_2-R_1)=\frac{5}{2}(2-3)=-2.5$

$\Rightarrow V_Q>V_P$