Question

$6\Omega$ and $12\Omega$ resistors are connected in parallel. This combination is connected in series with a $10V$ battery and $6\Omega$ resistor. What is the potential difference between the terminals of the $12\Omega$ resistor?

• A. 4 V
• B. 16 V
• C. 2
• D. 8 V

Answer 1

Answer: (A)

$R=\frac{6\times 12}{6+12}=\frac{6\times 12}{18}=4\Omega$
Total resistance,
$R_{eq}=6+4=10\Omega$
Current,$i=\frac{V}{R}=\frac{10}{10}=1A$
The current in $12\Omega$ resistor is
$i_2=i\left(\frac{R_1}{R_1+R_2}\right)=1\times \left(\frac{6}{6+12}\right)$
$i_2=\frac{1}{3}$
The potential difference in $12\Omega$ resistor
$V=i2R=\frac{1}{3}\times 12=4V$