Question
Find the current through the battery circuits shown in figure.
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Answer
Both diodes are forward biased. Thus the net diode resistance is 0. 
$\text{i}=\frac{5}{\frac{10\times10}{10+10}}=\frac{5}{5}=1\text{A}$
One diode is forward biased and other is reverse biased.
$\text{i}=\frac{\text{V}}{\text{R}_\text{net}}=\frac{5}{10+0}=\frac{1}{2}=0.5\text{A}$
$\text{i}=\frac{5}{\frac{10\times10}{10+10}}=\frac{5}{5}=1\text{A}$One diode is forward biased and other is reverse biased.
$\text{i}=\frac{\text{V}}{\text{R}_\text{net}}=\frac{5}{10+0}=\frac{1}{2}=0.5\text{A}$Need a full question paper?
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