In the following circuit,find the current thr | vedclass

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(B) The given circuit is a Wheatstone bridge. Let the potential at $A$ be $0 \ V$ and at $B$ be $16 \ V$. Since the bridge is balanced (all resistance

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$2$

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(B) The given circuit is a Wheatstone bridge. Let the potential at $A$ be $0 \ V$ and at $B$ be $16 \ V$. Since the bridge is balanced (all resistances are equal to $R = 4 \ \Omega$),the potential at $C$ and $D$ will be equal. However,we can simplify the circuit by noting that the path $ACB$ consists of two resistors in series,each of $R = 4 \ \Omega$. The total resistance of the path $ACB$ is $R_{ACB} = R + R = 4 \ \Omega + 4 \ \Omega = 8 \ \Omega$. The potential difference across the path $ACB$ is the same as the battery voltage,which is $16 \ V$. Using Ohm's law,the current $I$ through the path $ACB$ is given by $I = \frac{V}{R_{ACB}} = \frac{16 \ V}{8 \ \Omega} = 2 \ A$.

In the following circuit,find the current through the path $ACB$ if each resistance $R = 4 \ \Omega$ and the battery voltage is $16 \ V$. (in $A$)

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