Question

Resistor of resistance of $6\Omega$ each are connected in the manner shown in figure. With the current $0.5\text{A}$, the potential difference $V_A-V_B$

• A. $7.2\text{V}$
• B. $3.6\text{V}$
• C. $3.0\text{V}$
• D. $6.0\text{V}$

Answer 1

The correct option is C $3.0\text{V}$

The three $6\Omega$ resistors are connected in the parallel hence their equivalent resistance will be,

$R_1=\frac{R}{3}=\frac{6}{3}=2\Omega$

Similarly for the second part, the equivalent resistance is,

$R_2=\frac{12\times 6}{12+6}=\frac{72}{18}=4\Omega$

The given circuit can be reduced to


Therefore the equivalent resistance between $A$ and $B$ is,

$R_{AB}=2+4=6\Omega$

The current always flow from high potential region to a low potential region i.e $V_A>V_B$

Applying ohm's law,

$\Rightarrow V_A-V_B=i{R}_{AB}$

$\Rightarrow V_A-V_B=0.5\times 6=3\text{V}$

Why this question ? $\underset{–}{\text{Tip}}:$ In such problems always try to simplify the given circuit to the finest level, and then think to apply the ohm's law between two given points in order to find the potential difference between them.