Question

The reading of the voltmeter connected across $20\Omega$ resistance is

• A. $15V$
• B. $10V$
• C. $5V$
• D. $22.5V$

Answer 1

The correct option is A $15V$

The current distributions among the resistors is shown in figure.
Here, $I_1=\frac{4.5}{9}=0.5A$
As resistor (9,3) and 6 are in parallel so potential across 9$+$ (potential across 3$=$ potential across 3)
$\Rightarrow 4.5+3I_1=6I_2$ or $4.5+3\left(0.5\right)=6I_2$
or $I_2=\frac{4.5+1.5}{6}=1A$
So, $I=I_1+I_2=0.5+1=1.5A$
Equivalent resistance between C and D is $R_{cd}=(1/10+1/12+1/15{)}^{-1}=4\Omega$
Potential across C and D is $V_{cd}=I{R}_{cd}=1.5\times 4=6V$
Potential across A and D is
$V_{ad}=V_{ab}+V_{bc}+V_{cd}=6I_2+2I+6=6\left(1\right)+2\left(1.5\right)+6=6+3+6=15V$
This will be the potential across $20\Omega$ as well as the voltmeter because $20\Omega$ is connected in parallel across total resistance across A and D.