In the circuit shown the cells A and B have negligible resistance. For V _{ A }=12; V , R

Q: In the circuit shown the cells $A$ and $B$ have negligible resistance. For $V _{ A }=12\; V , R _{1}=500\; \Omega$ and $R =100\; \Omega$ the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$

c In the first loop which contains $V _{ A }$, the current is given by: $I =\frac{ V _{ A }}{ R _{1}+ R }$ $I =\frac{12}{500+100}$ $I =0.02 A$ Since negative terminal of battery is considered at zero voltage. Now voltage at the intersection point between $R$ and $R_{1}$ is given by: $V= IR$ $V =0.02 \times 100$\ $V=2 V$ So for battery $V_{B}$ of electric potential $2 V$ there will be no current flowing through the galvanometer as voltage difference between these points is zero.

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

In the circuit shown the cells $A$ and $B$ have negligible resistance. For $V _{ A }=12\; V , R _{1}=500\; \Omega$ and $R =100\; \Omega$ the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$

A$12$
B$6$
C$2$
D$4$

AIPMT 2012, Medium

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

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