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, then what is the potential difference between the points $P$ and $Q ?$

• A. Zero
• B. $4 V \left(V_{P} > V_{Q}\right)$
• C. $4 V \left(V_{Q} > V_{P}\right)$
• D. $2.5 V \left(V_{Q} > V_{P}\right)$

Answer 1

Answer: (D)

$R_{ eq }=\frac{5}{2} \,\Omega$
$i=\frac{20}{\frac{5}{2}+1.5}=5 \,A$

Potential difference between $X$ and $P$,
$V_{X}-V_{P}=\left(\frac{5}{2}\right) \times 3=7.5\, V \ldots$(i)
$V_{X}-V_{Q}=\frac{5}{2} \times 2=5\, V \ldots$ (ii)
On solving (i) and (ii),
$V_{P}-V_{Q}=-2.5\, V, V_{Q}>V_{P}$.