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
JEE
STD 12 - 3. current electricity
Physics

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