Discussion Forum

1)

 In given circuit , all resistance are of 10 Ω current flowing through ammeter is 

.

Answer:

Option A

Explanation:

 An equivalent of the given network is as shown in figure.

 If R_p be the net resistance, then

 

$\frac{1}{R_{p}}=\frac{1}{10}+\left(\frac{1}{10+10}\right)+\left(\frac{1}{10+10}\right)+\frac{1}{10}$

$=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\frac{6}{20}=\frac{3}{10}$

$\therefore$    $R_{p}=\frac{10}{3}$Ω

 Hence, current f;owing  through ammeter is 

  $I=\frac{V}{R_{p}}=\frac{12}{(\frac{10}{3})}=3.6 A$