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What is the current I shown in the given circuit? |
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Answer» `(V )/(2R)` In loop4 weget ` R_1=(2 R xx2R )/(2R+ 2R) = R ` In loop3, weget` R_2= ((R+ R_1)xx 2 R)/(2 R+(R+R_1))=R ` In loop2 we get`R_3= ((R +R_2)xx 2 R)/( 2 R+(R + R_2))=R` In loop1, weget ` R_4= R +R_3= R+ R =2R ` ` thereforeI =(I_1)/( 8 )= (V //2 R) /(8 )= (V )/( 16 R )` |
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