In the circuit diagram, potential difference between points A and B is 200 V, the potential difference between points a and b when the switch S is open, is

Question

In the circuit diagram, potential difference between points A and B is 200 V, the potential difference between points a and b when the switch S is open, is (A) 100 V (B) (200/3) V (C) (100/3) V (D) 50 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/54dc45c50eee477190cc8d99e68b63bc-.png" /> VA-Va=(200 × 6/3+6) =(200 × 6/9) V =(400/3) V VA-Vb=(200 × 3/3+6) =(200 × 3/9) V =(200/3) V VA-Va-(VA-Vb) =Vb-Va=(200/3) V

Q. In the circuit diagram, potential difference between points $A$ and $B$ is $200 V$ , the potential difference between points $a$ and $b$ when the switch $S$ is open, is

100 V

$\frac{200}{3} V$

$\frac{100}{3} V$

50 V

$V_{A} - V_{a} = \frac{200 \times 6}{3 + 6}$
$= \frac{200 \times 6}{9} V = \frac{400}{3} V$
$V_{A} - V_{b} = \frac{200 \times 3}{3 + 6}$
$= \frac{200 \times 3}{9} V = \frac{200}{3} V$
$V_{A} - V_{a} - (V_{A} - V_{b})$
$= V_{b} - V_{a} = \frac{200}{3} V$

Q. In the circuit diagram, potential difference between points A and B is 200V, the potential difference between points a and b when the switch S is open, is

100 V

3200​V

3100​V

50 V

VA​−Va​=3+6200×6​ =9200×6​V=3400​V VA​−Vb​=3+6200×3​ =9200×3​V=3200​V VA​−Va​−(VA​−Vb​) =Vb​−Va​=3200​V

Q. In the circuit diagram, potential difference between points $A$ and $B$ is $200 V$ , the potential difference between points $a$ and $b$ when the switch $S$ is open, is

$\frac{200}{3} V$

$\frac{100}{3} V$

$V_{A} - V_{a} = \frac{200 \times 6}{3 + 6}$
$= \frac{200 \times 6}{9} V = \frac{400}{3} V$
$V_{A} - V_{b} = \frac{200 \times 3}{3 + 6}$
$= \frac{200 \times 3}{9} V = \frac{200}{3} V$
$V_{A} - V_{a} - (V_{A} - V_{b})$
$= V_{b} - V_{a} = \frac{200}{3} V$