Calculate the current shown by the ammeter A | vedclass

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(B) The circuit consists of two $5 \, \Omega$ resistors in parallel with a bridge of four $10 \, \Omega$ resistors.First, consider the bridge part. Th

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$0.2$

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(B) The circuit consists of two $5 \, \Omega$ resistors in parallel with a bridge of four $10 \, \Omega$ resistors.First, consider the bridge part. The four $10 \, \Omega$ resistors are connected in a way that they form two parallel branches, each containing two $10 \, \Omega$ resistors in series. Resistance of each branch = $10 \, \Omega + 10 \, \Omega = 20 \, \Omega$.Since there are two such branches in parallel, the equivalent resistance of the bridge part $(R_b)$ is:$\frac{1}{R_b} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \implies R_b = 10 \, \Omega$.Now, this $10 \, \Omega$ equivalent resistance is in parallel with the two $5 \, \Omega$ resistors.Total equivalent resistance $(R_{	ext{net}})$ is given by:$\frac{1}{R_{	ext{net}}} = \frac{1}{5} + \frac{1}{5} + \frac{1}{10} = \frac{2+2+1}{10} = \frac{5}{10} = \frac{1}{2} \implies R_{	ext{net}} = 2 \, \Omega$.Using Ohm's law, $V = i R_{	ext{net}}$:$0.4 = i 	imes 2$$i = \frac{0.4}{2} = 0.2 \, A$.

Calculate the current shown by the ammeter $A$ in the circuit diagram. (in $A$)

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