Consider the network shown below with $R_{1} = 1 Ω$ , $R_{2} = 2 Ω$ and $R_{3} = 3 Ω$ . The network is connected to a constant voltage source of 11 V.

The magnitude of the current (in amperes, accurate to two decimal places) through the sources is

8

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Solution

The correct option is A 8

As the network is symmetric,
$V_{A} = V_{B} a n d V_{C} = V_{D}$

So, current through $R_{2} r e s i s t o r s i s z e r o a n d a s V_{A} = V_{B} a n d V_{C} = V_{D}$ , electrically the circuit can be reduced as,
Total resistance,

$R_{T} = 2 (R_{1} | | R_{1}) + (R_{1} | | R_{1} | | R_{3} | | R_{3})$

$= R_{1} + (\frac{R_{1}}{2} | | \frac{R_{3}}{2})$

Given that, $R_{1} = 1 Ω a n d R_{3} = 3 Ω$

So, $R_{T} = 1 + (\frac{1}{2} | | \frac{3}{2}) Ω = 1 + \frac{3 / 2}{4} = \frac{11}{8} Ω$

$I = \frac{11 V}{R_{T}} = \frac{11}{(11 / 8)} = 8$

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