Question

$6 \,\Omega$ and $12 \,\Omega$ resistors are connected in parallel. This combination is connected in series with a $10\, V$ 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,$i = \frac{V}{R} = \frac{10}{10} = 1\,A$
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
$V = i\,2R = \frac{1}{3} \times 12 = 4\,V$