Four resistances $40\Omega,\ 60\Omega,\ 90\Omega\ and\ 110\Omega$ make the arms of a quadrilateral $ABCD$. Across $AC$ is the battery circuit, the emf of the battery being 4V and internal resistance negligible. The potential difference across $BD$ is

Question

Dr. Abubacker · Accepted Answer

(d) Current through arm $ABC$,
$= 4/(40 + 60) = 0.04\ A$
Potential difference across $A$ and $B$,
$V_{A} - V_{B} = 0.04 	imes 40 = 1.6V$
Current through arm $ADC$,
$= 4/(90 + 110) = 4/200A$
Potential difference between $A$ and $D$,
$V_{A} - V_{D} = \frac{4}{200} 	imes 90 = 1.8V$
$	herefore V_{B} - V_{D} = \left( V_{A} - V_{D} ight) - (V_{A} - V_{B})$
$1.8 - 1.6 = 0.2\ V$

Q. Four resistances $40\Omega,\ 60\Omega,\ 90\Omega\ and\ 110\Omega$ make the arms of a quadrilateral $ABCD$. Across $AC$ is the battery circuit, the emf of the battery being 4V and internal resistance negligible. The potential difference across $BD$ is

Explanation

Reviewed by Dr. Abubacker

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