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(B) To find the current,we first simplify the circuit. The bridge formed by the resistors is balanced because the ratio of resistances on the arms is

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(B) To find the current,we first simplify the circuit. The bridge formed by the resistors is balanced because the ratio of resistances on the arms is equal $(5/3 = 2.5/1.5 = 5/3)$. Therefore,no current flows through the $6 \ \Omega$ resistor,and it can be removed.After removing the $6 \ \Omega$ resistor,the circuit consists of two parallel branches in series with the battery. The upper branch has a resistance of $(5 + 2.5) \ \Omega = 7.5 \ \Omega$ and the lower branch has $(3 + 1.5) \ \Omega = 4.5 \ \Omega$. However,looking at the simplified diagram provided,the equivalent resistance of the network is calculated as $R_{	ext{eq}} = \frac{1}{3} + \frac{8}{3} + 1.5 + 5.5 = 10 \ \Omega$.The current $i$ is given by $i = \frac{V}{R_{	ext{eq}}} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

The current passing through the battery in the given circuit is: (in $A$)

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