A battery of e.m.f. 10, V is connected to resistance as shown in figure. The potential difference {V_A}

Q: A battery of $e.m.f.$ $10\, V$ is connected to resistance as shown in figure. The potential difference ${V_A} - {V_B}$ between the points $A$ and $B$ is .................... $V$

b (b) ${R_{eq}} = 5\,\Omega $, Current $i = \frac{{10}}{5} = 2\,A$ and current in each branch = $1\,A$ Potential difference between $C$ and $A$, ${V_C} - {V_A} = 1 \times 1 = 1\,V$ .......$(i)$ Potential difference between $C$ and $B$, ${V_C} - {V_B} = 1 \times 3 = 3\,V$......$(ii)$ On solving $(i)$ and $(ii)$ ${V_A} - {V_B} = 2\,volt$ Shot Trick : $({V_A} - {V_B}) = \frac{i}{2}({R_2} - {R_1}) = \frac{2}{2}(3 - 1) = 2\,V$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A battery of $e.m.f.$ $10\, V$ is connected to resistance as shown in figure. The potential difference ${V_A} - {V_B}$ between the points $A$ and $B$ is .................... $V$

A$ - 2$
B$2$
C$5$
D$\frac{{20}}{{11}}$

Diffcult

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Download our appand get started for free

Similar Questions

Download our app
and get started for free