Question 109 of 380

In the given circuit, the current $I$ through the battery will be

A$2.5A$B$1A$C$2A$D$1.5A$

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Answer: (D) $1.5A$

In the first circuit diagram, the diodes {D}_{1}&{D}_{3} are forward biased and ${D}_{2}$ is reverse biased(Hence, no current through it). So the circuit is simplified to the second diagram.
Hence, the equivalent resistance is
${R}_{eq}=\dfrac{20\times 10}{20+10}$
$\Rightarrow {R}_{eq}=\dfrac{20}{3}\Omega$
Using Ohm's law, the current is
$I=\dfrac{10}{\dfrac{20}{3}}=\dfrac{30}{20}A=1.5A$

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