Question # Find the ratio of currents as measured by ideal ammeter in two cases when the key ( ( k ) ) is open and when the key ( ( k ) ) is closed, respectively.

# Find the ratio of currents as measured by ideal ammeter in two cases when the key ( ( k ) ) is open and when the key ( ( k ) ) is closed, respectively.

( begin{array}{ll}text { A } & text { 9/11 }end{array} )

( begin{array}{ll}text { B } & 10 / 11end{array} )

C(quad 8 / 9 )

D ( 8 / 11 )

Solution

When key is open

[

begin{array}{l}

operatorname{Reg}=frac{(2 R+R)(R+2 R)}{(2 R+R)+(R+2 R)}=frac{9 R^{2}}{6 R}=frac{3 R}{2}

i=frac{2 E}{3 R}

end{array}

]

When key is dosed.

[

]

[

begin{array}{l}

r_{p}=frac{R times 2 R}{R+2 R}+frac{2 R times R}{R+2 R}=frac{4 R}{3}

I_{2}= frac{3 E}{4 R} Rightarrow frac{l_{1}}{l_{2}}=frac{2 E / 3 R}{3 E / 4 R}=frac{8}{9}

end{array}

]

option C