Question

IN THE SHOWN NETWORK THE CURRENT I IN THE CELL SEGMENT IS​

Answer 1

Answer:

$\sf{the \: above \: arrangement \: is \: forming \: a} \\ \sf{wheatstone \: bridge \: so \: the \: resistance \: in} \\ \sf{mid \: wire \: i.e. \: 5 \: ohm \: can \: be \: neglected.}$

$\star \: now \: 5 \: and \: 15 \: ohm \:resistance \: in \: upper \: \\ wire\: are \: in \: series \rightarrow \\ \\ \sf{total \: resistance \: = 5 + 15 = 20 \: ohm} \\ \\ \star \: same \: will \: be \: for \: the \: lower \: wire \\ \\ \sf{total \: resistance \: = \: 20 \: ohm}$

$\bf{these \: two \: 20 \: ohm \: resistances \: will} \\ \bf{ be \: parallel} \\ \\ \sf{ \frac{1}{R} = \frac{1}{20} + \frac{1}{20} } \\ \\ \sf{R \: = 10 \: ohm}$

$\bf{ \bullet \: the \:two \: 10 \: ohm \: resistances \: will \: be } \\ \: \: \: \bf{in \: series} \\ \\ \sf{net \: resistance \:R_{net}= 10 + 10 = 20 \: ohm} \\ \\ \sf{now \: current} \\ \boxed{I \: = \frac{V}{R_{net}} } \\ \\ I \: = \frac{10}{20} \\ \\ \bf \boxed{I \: = 0.5 \: ampere}$

$\mathscr{\large{Thanks}}\heartsuit$

Answer 2

Answer:

Answer is 0.2 A

Because this Anwer is given in Brilliant guess paper