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$6 \Omega$ and $12 \Omega$ resistors are connected in parallel. This combination is connected in series with a $10 \mathrm{~V}$ battery and $6 \Omega$ resistor. What is the potential difference between the terminals of the $12 \Omega$ resistor?
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The correct answer is:
$4 \mathrm{~V}$
$R=\frac{6 \times 12}{6+12}=\frac{6 \times 12}{18}=4 \Omega$
Total resistance,
$R_{\mathrm{eq}}=6+4=10 \Omega$
Current, $\quad i=\frac{V}{R}=\frac{10}{10}=1 \mathrm{~A}$
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
$V=i R=\frac{1}{3} \times 12=4 \mathrm{~V}$
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