Q134 / 160 · AP EAMCET 2020 (22 Sep Shift 1) · Physics
Potential difference between the points $P$ and $Q$ in the circuit shown is
- A$4.5 \mathrm{~V}$
- B$2.4 \mathrm{~V}$
- C2.4V
- Correct answerD2.88V
Solution
The given circuit diagram is
Equivalent resistance between point $P$ and $Q$ is given as $$ \begin{aligned} \frac{1}{R_{P Q}} & =\frac{1}{R_A+R_D}+\frac{1}{3}+\frac{1}{R_B+R_C} \\ & =\frac{1}{2+6}+\frac{1}{3}+\frac{1}{4+12}=\frac{1}{8}+\frac{1}{3}+\frac{1}{16} \\ \Rightarrow \quad \frac{1}{R_{P Q}} & =\frac{25}{48} \Rightarrow R_{P Q}=\frac{48}{25} \Omega \end{aligned} $$ $\therefore$ Potential difference between the points $P$ and $Q$ is given as $$ V_{P Q}=I \cdot R_{P Q}=1.5 \times \frac{48}{25}=2.88 \mathrm{~V} $$
Equivalent resistance between point $P$ and $Q$ is given as $$ \begin{aligned} \frac{1}{R_{P Q}} & =\frac{1}{R_A+R_D}+\frac{1}{3}+\frac{1}{R_B+R_C} \\ & =\frac{1}{2+6}+\frac{1}{3}+\frac{1}{4+12}=\frac{1}{8}+\frac{1}{3}+\frac{1}{16} \\ \Rightarrow \quad \frac{1}{R_{P Q}} & =\frac{25}{48} \Rightarrow R_{P Q}=\frac{48}{25} \Omega \end{aligned} $$ $\therefore$ Potential difference between the points $P$ and $Q$ is given as $$ V_{P Q}=I \cdot R_{P Q}=1.5 \times \frac{48}{25}=2.88 \mathrm{~V} $$
Hence, the correct answer is option: 'D'.