Find the ratio of currents as measu...
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.
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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 )

JEE/Engineering Exams
Physics
Solution
94
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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.
Fullscreen

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

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